1 Million+ Step-by-step solutions

Two couples act on the frame. Determine the resultant couple moment. Compute the result by resolving each force into x and y components and

(a) Finding the moment of each couple (Eq. 4 -13) and

(b) Summing the moments of all the force components about point B.

Given:

F1 = 80 lb d = 4ft

F2 = 50 lb e = 3 ft

a = 1 ft f = 3 ft

b = 3 ft g = 4 ft

c = 2 ft θ = 30 deg

(a) Finding the moment of each couple (Eq. 4 -13) and

(b) Summing the moments of all the force components about point B.

Given:

F1 = 80 lb d = 4ft

F2 = 50 lb e = 3 ft

a = 1 ft f = 3 ft

b = 3 ft g = 4 ft

c = 2 ft θ = 30 deg

Determine the couple moment. Express the result as a Cartesian vector. Determine the couple moment. Express the result as a Cartesian vector.

Given:

F = (8 -4 10)) N

a = 5 m

b = 3 m

c = 4 m

d = 2 m

e = 3 m

Given:

F = (8 -4 10)) N

a = 5 m

b = 3 m

c = 4 m

d = 2 m

e = 3 m

Determine the couple moment. Express the result as a Cartesian vector.

Given:

F = 80 N

a = 6 m

b = 10 m

c = 10 m

d = 5 m

e = 4 m

f = 4 m

Given:

F = 80 N

a = 6 m

b = 10 m

c = 10 m

d = 5 m

e = 4 m

f = 4 m

If the resultant couple of the two couples acting on the fire hydrant is MR = {−15i + 30j} N ∙ m, determine the force magnitude P.

Given:

a = 0.2 m

b = 0.150 m

M = (-15 30 0) N ∙ m

F = 75 N

Given:

a = 0.2 m

b = 0.150 m

M = (-15 30 0) N ∙ m

F = 75 N

If the resultant couple of the three couples acting on the triangular block is to be zero, determine the magnitude of forces F and P.

Given:

F1 = 150 N

a = 300 mm

b = 400 mm

d = 600 mm

Given:

F1 = 150 N

a = 300 mm

b = 400 mm

d = 600 mm

Determine the couple moment that acts on the assembly. Express the result as a Cartesian vector. Member BA lies in the x-y plane.

Given:

F = (0 0 100) N

a = 300 mm

b = 150 mm

c = 200 mm

d = 200 mm

θ = 60 deg

Given:

F = (0 0 100) N

a = 300 mm

b = 150 mm

c = 200 mm

d = 200 mm

θ = 60 deg

If the magnitude of the resultant couple moment is M, determine the magnitude F of the forces applied to the wrenches.

Given:

M = 15 N ∙ mc = 200 mm

a = 300 mm d = 200 mm

b = 150 mm θ = 60 deg

Given:

M = 15 N ∙ mc = 200 mm

a = 300 mm d = 200 mm

b = 150 mm θ = 60 deg

The gears are subjected to the couple moments shown. Determine the magnitude and coordinate direction angles of the resultant couple moment.

Given:

M1 = 40 lb⋅ ft

M2 = 30 lb⋅ ft

θ1 = 20 deg

θ2 = 15 deg

θ3 = 30 deg

Given:

M1 = 40 lb⋅ ft

M2 = 30 lb⋅ ft

θ1 = 20 deg

θ2 = 15 deg

θ3 = 30 deg

Express the moment of the couple acting on the rod in Cartesian vector form. What is the magnitude of the couple moment?

Given:

a = 1.5 m

b = 0.5 m

c = 0.5 m

d = 0.8 m

Given:

a = 1.5 m

b = 0.5 m

c = 0.5 m

d = 0.8 m

Express the moment of the couple acting on the pipe assembly in Cartesian vector form. Solve the problem

(a) Using Eq. 4-13,

(b) Summing the moment of each force about point O.

Given:

a = 0.3 m

b = 0.4 m

c = 0.6 m

F = (- 6 2 3) N

A couple acts on each of the handles of the minidual valve. Determine the magnitude and coordinate direction angles of the resultant couple moment.

Given:

F1 = 35 N θ = 60 deg

F2 = 25 N

r1 = 175 mm

r2 = 175 mm

Given:

F1 = 35 N θ = 60 deg

F2 = 25 N

r1 = 175 mm

r2 = 175 mm

Express the moment of the couple acting on the pipe in Cartesian vector form. What is the magnitude of the couple moment?

Given:

F = 125 N

a = 150 mm

b = 150 mm

c = 200 mm

d = 600 mm

Given:

F = 125 N

a = 150 mm

b = 150 mm

c = 200 mm

d = 600 mm

If the couple moment acting on the pipe has a magnitude M, determine the magnitude F of the forces applied to the wrenches.

Given:

M 300 = N ∙ m

a = 150 mm

b = 150 mm

c = 200 mm

d = 600 mm

Given:

M 300 = N ∙ m

a = 150 mm

b = 150 mm

c = 200 mm

d = 600 mm

Replace the force at A by an equivalent force and couple moment at point O.

Given:

F = 375 N

a = 2 m

b = 4 m

c = 2 m

d = 1 m

θ = 30 deg

Given:

F = 375 N

a = 2 m

b = 4 m

c = 2 m

d = 1 m

θ = 30 deg

Replace the force at A by an equivalent force and couple moment at point P.

Given:

F = 375 N

a = 2 m

b = 4 m

c = 2 m

d = 1 m

θ = 30 deg

Given:

F = 375 N

a = 2 m

b = 4 m

c = 2 m

d = 1 m

θ = 30 deg

Replace the force system by an equivalent resultant force and couple moment at point O.

Given:

F1 = 60 lb a = 2 ft

F2 = 85 lb b = 3 ft

F3 = 25 lb c = 6 ft

θ = 45 deg d = 4 ft

e = 3

f = 4

Given:

F1 = 60 lb a = 2 ft

F2 = 85 lb b = 3 ft

F3 = 25 lb c = 6 ft

θ = 45 deg d = 4 ft

e = 3

f = 4

Replace the force system by an equivalent resultant force and couple moment at point P.

Given:

F1 = 60 lb a = 2 ft

F2 = 85 lb b = 3 ft

F3 = 25 lb c = 6 ft

θ = 45 deg d = 4 ft

e = 3 f = 4

Given:

F1 = 60 lb a = 2 ft

F2 = 85 lb b = 3 ft

F3 = 25 lb c = 6 ft

θ = 45 deg d = 4 ft

e = 3 f = 4

Replace the force system by an equivalent force and couple moment at point O.

Units Used:

kip = 103 lb

Given:

F1 = 430 lb F2 = 260 lb

a = 2 ft e = 5 ft

= 8 ft f = 12

c = 3 ft g = 5

d =a θ = 60 deg

Units Used:

kip = 103 lb

Given:

F1 = 430 lb F2 = 260 lb

a = 2 ft e = 5 ft

= 8 ft f = 12

c = 3 ft g = 5

d =a θ = 60 deg

Replace the force system by an equivalent force and couple moment at point P.

Units Used:

kip = 103 lb

Given:

F1 = 430 lb F2 = 260 lb

a = 2 ft e = 5 ft

b = 8 ft f = 12

c = 3 ft g = 5

d =a θ = 60 deg

Units Used:

kip = 103 lb

Given:

F1 = 430 lb F2 = 260 lb

a = 2 ft e = 5 ft

b = 8 ft f = 12

c = 3 ft g = 5

d =a θ = 60 deg

Replace the loading system acting on the post by an equivalent resultant force and couple moment at point O.

Given:

F1 = 30 lb a = 1 ft d = 3

F2 = 40 lb b = 3 ft e = 4

F3 = 60 lb c = 2 ft

Given:

F1 = 30 lb a = 1 ft d = 3

F2 = 40 lb b = 3 ft e = 4

F3 = 60 lb c = 2 ft

Replace the loading system acting on the post by an equivalent resultant force and couple moment at point P.

Given:

F1 = 30 lb

F2 = 40 lb

F3 = 60 lb

a = 1 ft

b = 3 ft

c = 2 ft

d = 3

e = 4

Given:

F1 = 30 lb

F2 = 40 lb

F3 = 60 lb

a = 1 ft

b = 3 ft

c = 2 ft

d = 3

e = 4

Replace the force and couple system by an equivalent force and couple moment at point O.

Units Used:

kN = 103 N

Given:

M = 8kNm θ = 60 deg

a = 3 m f = 12

b = 3 m g = 5

c = 4 m F1 = 6kN

d = 4m F2 = 4kN

e = 5 m

Replace the force and couple system by an equivalent force and couple moment at point P.

Units Used:

kN = 103 N

Given:

M 8kN = ⋅m θ = 60 deg

a = 3 m f = 12

b = 3 m g = 5

c = 4 m F1 = 6kN

d = 4 m F2 = 4kN

e = 5 m

Replace the force system by a single force resultant and specify its point of application, measured along the x axis from point O.

Given:

F1 = 125 lb

F2 = 350 lb

F3 = 850 lb

a = 2 ft

b = 6 ft

c = 3 ft

d = 4 ft

Given:

F1 = 125 lb

F2 = 350 lb

F3 = 850 lb

a = 2 ft

b = 6 ft

c = 3 ft

d = 4 ft

Replace the force system by a single force resultant and specify its point of application, measured along the x axis from point P.

Given:

F1 = 125 lb a = 2 ft

F2 = 350 lb b = 6 ft

F3 = 850 lb c = 3 ft

d = 4 ft

Given:

F1 = 125 lb a = 2 ft

F2 = 350 lb b = 6 ft

F3 = 850 lb c = 3 ft

d = 4 ft

The forces and couple moments which are exerted on the toe and heel plates of a snow ski are Ft, Mt, and Fh, Mh, respectively. Replace this system by an equivalent force and couple moment acting at point O. Express the results in Cartesian vector form.

Given:

a = 120 mm

b = 800 mm

The forces and couple moments which are exerted on the toe and heel plates of a snow ski are Ft, Mt, and Fh, Mh, respectively. Replace this system by an equivalent force and couple moment acting at point P. Express the results in Cartesian vector form.

Given:

a = 120 mm

b = 800 mm

Ft = (-50 80 -158) N

Mt = (-6 4 2) N ∙ m

Fh = (-20 60 -250) N

Mh = (-20 8 3) N. m

Replace the three forces acting on the shaft by a single resultant force. Specify where the force acts, measured from end B.

Given:

F1 = 500 lb

F2 = 200 lb

F3 = 260 lb

a = 5 ft e = 3

b = 3 ft f = 4

c = 2 ft g = 12

d = 4 ft h = 5

Given:

F1 = 500 lb

F2 = 200 lb

F3 = 260 lb

a = 5 ft e = 3

b = 3 ft f = 4

c = 2 ft g = 12

d = 4 ft h = 5

Given:

F1 = 500 lb

F2 = 200 lb

F3 = 260 lb

a = 5 ft e = 3

b = 3 ft f = 4

c = 2 ft g = 12

d = 4 ft h = 5

Replace the loading on the frame by a single resultant force. Specify where its line of action intersects member AB, measured from A.

Given:

F1 = 300 lb M = 600 lb⋅ ft

F2 = 200 lb a = 3 ft

b = 4 ft

F3 = 400 lb

c = 2 ft

F4 = 200 lb d = 7 ft

Given:

F1 = 300 lb M = 600 lb⋅ ft

F2 = 200 lb a = 3 ft

b = 4 ft

F3 = 400 lb

c = 2 ft

F4 = 200 lb d = 7 ft

Replace the loading on the frame by a single resultant force. Specify where the force acts, measured from end A.

Given:

F1 = 450 N a = 2 m

F2 = 300 N b = 4 m

F3 = 700 N c = 3 m

θ = 60 deg M = 1500 N ∙ m

φ = 30 deg

Given:

F1 = 450 N a = 2 m

F2 = 300 N b = 4 m

F3 = 700 N c = 3 m

θ = 60 deg M = 1500 N ∙ m

φ = 30 deg

Replace the loading on the frame by a single resultant force. Specify where the force acts , measured from end B.

Given:

F1 = 450 N a = 2 m

F2 = 300 N b = 4 m

F3 = 700 N c = 3 m

θ = 60 deg M = 1500 N ∙ m

φ = 30deg

Given:

F1 = 450 N a = 2 m

F2 = 300 N b = 4 m

F3 = 700 N c = 3 m

θ = 60 deg M = 1500 N ∙ m

φ = 30deg

Replace the loading system acting on the beam by an equivalent resultant force and couple moment at point O.

Given:

F1 = 200 N

F2 = 450 N

M = 200N ∙ m

a = 0.2 m

b = 1.5 m

c = 2 m

d = 1.5 m

θ = 30 deg

Given:

F1 = 200 N

F2 = 450 N

M = 200N ∙ m

a = 0.2 m

b = 1.5 m

c = 2 m

d = 1.5 m

θ = 30 deg

Determine the magnitude and direction θ of force F and its placement d on the beam so that the loading system is equivalent to a resultant force FR acting vertically downward at point A and a clockwise couple moment M.

Units Used:

kN = 103 N

Given:

F1 = 5kN a = 3 m

F2 = 3kN b = 4 m

FR = 12kN c = 6 m

M = 50kN ∙ m e = 7 f = 24

Units Used:

kN = 103 N

Given:

F1 = 5kN a = 3 m

F2 = 3kN b = 4 m

FR = 12kN c = 6 m

M = 50kN ∙ m e = 7 f = 24

Determine the magnitude and direction θ of force F and its placement d on the beam so that the loading system is equivalent to a resultant force FR acting vertically downward at point A and a clockwise couple moment M.

Units Used:

kN = 103 N

Given:

F1 = 5kN a = 3 m

F2 = 3kN b = 4 m

FR = 10kN c = 6 m

M = 45kN ∙ m e = 7

f = 24

Units Used:

kN = 103 N

Given:

F1 = 5kN a = 3 m

F2 = 3kN b = 4 m

FR = 10kN c = 6 m

M = 45kN ∙ m e = 7

f = 24

Replace the loading on the frame by a single resultant force. Specify where its line of action intersects member AB, measured from A.

Given:

F1 = 500 N a = 3 m

F2 = 300 N b = 2 m

c = 1 m

F3 = 250 N

d = 2 m

M = 400 N ∙ m e = 3 m

θ = 60 deg f = 3

g = 4

Replace the loading on the frame by a single resultant force. Specify where its line of action intersects member CD, measured from end C.

Given:

F1 = 500 N a = 3 m

F2 = 300 N b = 2 m

c = 1 m

F3 = 250 N

d = 2 m

M = 400 N ∙ m e = 3 m

θ = 60 deg f = 3

g = 4

Given:

F1 = 500 N a = 3 m

F2 = 300 N b = 2 m

c = 1 m

F3 = 250 N

d = 2 m

M = 400 N ∙ m e = 3 m

θ = 60 deg f = 3

g = 4

Replace the force system acting on the frame by an equivalent resultant force and specify where the resultant's line of action intersects member AB, measured from point A.

Given:

F1 = 35 lb a = 2 ft

F2 = 20 lb b = 4 ft

F3 = 25 lb c = 3 ft

θ = 30 deg d = 2 ft

Given:

F1 = 35 lb a = 2 ft

F2 = 20 lb b = 4 ft

F3 = 25 lb c = 3 ft

θ = 30 deg d = 2 ft

Replace the force system acting on the frame by an equivalent resultant force and specify where the resultant's line of action intersects member BC, measured from point B.

Given:

F1 = 35 lb

F2 = 20 lb

F3 = 25 lb

θ = 30 deg

a = 2 ft

b = 4 ft

c = 3 ft

d = 2 ft

Given:

F1 = 35 lb

F2 = 20 lb

F3 = 25 lb

θ = 30 deg

a = 2 ft

b = 4 ft

c = 3 ft

d = 2 ft

Replace the force system acting on the frame by an equivalent resultant force and couple moment acting at point A.

Given:

F1 = 35 lb a = 2 ft

F2 = 20 lb b = 4 ft

F3 = 25 lb c = 3 ft

θ = 30 deg d = 2 ft

Given:

F1 = 35 lb a = 2 ft

F2 = 20 lb b = 4 ft

F3 = 25 lb c = 3 ft

θ = 30 deg d = 2 ft

Replace the force and couple-moment system by an equivalent resultant force and couple moment at point O. Express the results in Cartesian vector form.

Units Used:

kN = 103 N

Units Used:

kN = 103 N

Replace the force and couple-moment system by an equivalent resultant force and couple moment at point O. Express the results in Cartesian vector form.

Units Used:

kN = 103 N

Given:

F = (8 6 8) kN

M = (-20 -70 20) kN ∙ m

a = 3 m

b = 3 m e = 5 m

c = 4 m f = 6 m

d = 6 m g = 5 m

Units Used:

kN = 103 N

Given:

F = (8 6 8) kN

M = (-20 -70 20) kN ∙ m

a = 3 m

b = 3 m e = 5 m

c = 4 m f = 6 m

d = 6 m g = 5 m

Replace the force and couple-moment system by an equivalent resultant force and couple moment a

t point Q. Express the results in Cartesian vector form.

Units Used:

kN = 103 N

Given:

F = (8 6 8) kN

M = (-20 -70 20) k N ∙ m

a = 3 m

b = 3 m e = 5 m

c = 4 m f = 6 m

d = 6 m g = 5 m

t point Q. Express the results in Cartesian vector form.

Units Used:

kN = 103 N

Given:

F = (8 6 8) kN

M = (-20 -70 20) k N ∙ m

a = 3 m

b = 3 m e = 5 m

c = 4 m f = 6 m

d = 6 m g = 5 m

The belt passing over the pulley is subjected to forces F1 and F2. F1 acts in the k direction. Replace these forces by an equivalent force and couple moment at point A. Express the result in Cartesian vector form.

Given:

F1 = 40 N r = 80 mm

F2 = 40 N a = 300 mm

θ = 0 deg

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Given:

F1 = 40 N r = 80 mm

F2 = 40 N a = 300 mm

θ = 0 deg

The belt passing over the pulley is subjected to forces F1 and F2. F1 acts in the −k direction. Replace these forces by an equivalent force and couple moment at point A. Express the result in Cartesian vector form.

Given:

F1 = 40 N

F2 = 40 N

θ = 0 deg

r = 80 mm

a = 300 mm

θ = 45 deg

Given:

F1 = 40 N

F2 = 40 N

θ = 0 deg

r = 80 mm

a = 300 mm

θ = 45 deg

Replace this system by an equivalent resultant force and couple moment acting at O. Express the results in Cartesian vector form.

Given:

F1 = 50 N

F2 = 80 N

F3 = 180 N

a = 1.25 m

b = 0.5 m

c = 0.75 m

Given:

F1 = 50 N

F2 = 80 N

F3 = 180 N

a = 1.25 m

b = 0.5 m

c = 0.75 m

Handle forces F1 and F2 are applied to the electric drill. Replace this system by an equivalent resultant force and couple moment acting at point O. Express the results in Cartesian vector form.

Given:

a = 0.15 m

b = 0.25 m

c = 0.3 m

F1 = (6 -3 -10) N

F2 = (0 2 -4) N

Given:

a = 0.15 m

b = 0.25 m

c = 0.3 m

F1 = (6 -3 -10) N

F2 = (0 2 -4) N

A biomechanical model of the lumbar region of the human trunk is shown. The forces acting in the four muscle groups consist of FR for the rectus, FO for the oblique, FL for the lumbar latissimus dorsi, and FE for the erector spinae. These loadings are symmetric with respect to the y - z plane. Replace this system of parallel forces by an equivalent force and couple moment acting at the spine, point O. Express the results in Cartesian vector form.

Given:

FR = 35 N a = 75 mm

FO = 45 N b = 45 mm

FL = 23 N c = 15 mm

FE = 32 N d = 50 mm

e = 40 mm f = 30 mm

Given:

FR = 35 N a = 75 mm

FO = 45 N b = 45 mm

FL = 23 N c = 15 mm

FE = 32 N d = 50 mm

e = 40 mm f = 30 mm

The building slab is subjected to four parallel column loadings. Determine the equivalent resultant force and specify its location (x, y) on the slab.

Units Used:

kN = 103 N

Given:

F1 = 30 kN a = 3 m

F2 = 40 kN b = 8 m

F3 = 20 kN c = 2 m

F4 = 50 kN d = 6 m

e = 4 m

Units Used:

kN = 103 N

Given:

F1 = 30 kN a = 3 m

F2 = 40 kN b = 8 m

F3 = 20 kN c = 2 m

F4 = 50 kN d = 6 m

e = 4 m

The building slab is subjected to four parallel column loadings. Determine the equivalent resultant force and specify its location (x, y) on the slab.

Units Used:

kN = 103 N

Given:

F1 = 20 kN a = 3 m

F2 = 50 kN b = 8 m

F3 = 20 kN c = 2 m

F4 = 50 kN d = 6 m

e = 4 m

Units Used:

kN = 103 N

Given:

F1 = 20 kN a = 3 m

F2 = 50 kN b = 8 m

F3 = 20 kN c = 2 m

F4 = 50 kN d = 6 m

e = 4 m

The pipe assembly is subjected to the action of a wrench at B and a couple at A. Determine the magnitude F of the couple forces so that the system can be simplified to a wrench acting at point C.

Given:

a = 0.6 m

b = 0.8 m

c = 0.25 m

d = 0.7 m

e = 0.3 m

f = 0.3 m

g = 0.5 m

h = 0.25 m

P = 60 N

Q = 40 N

Given:

a = 0.6 m

b = 0.8 m

c = 0.25 m

d = 0.7 m

e = 0.3 m

f = 0.3 m

g = 0.5 m

h = 0.25 m

P = 60 N

Q = 40 N

The three forces acting on the block each have a magnitude F1 = F2 = F3. Replace this system by a wrench and specify the point where the wrench intersects the z axis, measured from point O.

Given:

F1 = 10 lb a = 6 ft

F2 = F1 b = 6 ft

F3 = F1 c = 2 ft

Given:

F1 = 10 lb a = 6 ft

F2 = F1 b = 6 ft

F3 = F1 c = 2 ft

Replace the three forces acting on the plate by a wrench. Specify the magnitude of the force and couple moment for the wrench and the point P(x, y) where its line of action intersects the plate.

Units Used:

kN = 103N

Given:

FA = 500 N

FB = 800 N

FC = 300 N

a = 4 m

b = 6 m

Units Used:

kN = 103N

Given:

FA = 500 N

FB = 800 N

FC = 300 N

a = 4 m

b = 6 m

Replace the three forces acting on the plate by a wrench. Specify the magnitude of the force and couple moment for the wrench and the point P(y, z) where its line of action intersects the plate.

Given:

FA = 80 lb a = 12 ft

FB = 60 lb b = 12 ft

FC = 40 lb

Given:

FA = 80 lb a = 12 ft

FB = 60 lb b = 12 ft

FC = 40 lb

The loading on the bookshelf is distributed as shown. Determine the magnitude of the equivalent resultant location, measured from point O.

Given:

w1 = 2lb/ft

w2 = 3.5lb/ft

a = 2.75 ft

b = 4 ft

c = 1.5 ft

Given:

w1 = 2lb/ft

w2 = 3.5lb/ft

a = 2.75 ft

b = 4 ft

c = 1.5 ft

Replace the loading by an equivalent resultant force and couple moment acting at point A.

Units Used:

kN = 103 N

Given:

w1 600N/m

w2 = 600 N/m

a = 2.5 m

b = 2.5 m

Units Used:

kN = 103 N

Given:

w1 600N/m

w2 = 600 N/m

a = 2.5 m

b = 2.5 m

Replace the loading by an equivalent force and couple moment acting at point O.

Units Used:

kN = 103 N

Given:

w = 6 kN/m

F = 15 kN

M = 500 kN ∙ m

a = 7.5 m

b = 4.5 m

Units Used:

kN = 103 N

Given:

w = 6 kN/m

F = 15 kN

M = 500 kN ∙ m

a = 7.5 m

b = 4.5 m

Replace the loading by a single resultant force, and specify the location of the force on the beam measured from point O.

Units Used:

kN = 103 N

Given:

w = 6 kN/m

F = 15 kN

M = 500 kN ∙ m

a = 7.5 m

b = 4.5 m

Units Used:

kN = 103 N

Given:

w = 6 kN/m

F = 15 kN

M = 500 kN ∙ m

a = 7.5 m

b = 4.5 m

The column is used to support the floor which exerts a force P on the top of the column.

The effect of soil pressure along its side is distributed as shown. Replace this loading by an equivalent resultant force and specify where it acts along the column, measured from its base A.

Units Used: kip = 103 lb

Given:

P = 3000 lb

w1 = 80lb/ft

w2 = 200 lb/ft

h = 9 ft

The effect of soil pressure along its side is distributed as shown. Replace this loading by an equivalent resultant force and specify where it acts along the column, measured from its base A.

Units Used: kip = 103 lb

Given:

P = 3000 lb

w1 = 80lb/ft

w2 = 200 lb/ft

h = 9 ft

Replace the loading by an equivalent resultant force and specify its location on the beam, measured from point B.

Units Used:

kip = 103 lb

Given:

w1 = 800lb/ft

w2 = 500lb/ft

a = 12 ft

b = 9 ft

Units Used:

kip = 103 lb

Given:

w1 = 800lb/ft

w2 = 500lb/ft

a = 12 ft

b = 9 ft

Replace the distributed loading by an equivalent resultant force, and specify its location on the beam, measured from the pin at C.

Units Used:

kip = 103 lb

Given:

w = 800 lb/ft

a = 15 ft

b = 15 ft

θ = 30 deg

Units Used:

kip = 103 lb

Given:

w = 800 lb/ft

a = 15 ft

b = 15 ft

θ = 30 deg

The beam supports the distributed load caused by the sandbags. Determine the resultant force on the beam and specify its location measured from point A.

Units Used: kN = 103 N

Given:

w1 = 1.5 kN/m a = 3 m

w2 = 1kN/m b = 3 m

w3 = 2.5kN/m c = 1.5 m

Units Used: kN = 103 N

Given:

w1 = 1.5 kN/m a = 3 m

w2 = 1kN/m b = 3 m

w3 = 2.5kN/m c = 1.5 m

Determine the length b of the triangular load and its position a on the beam such that the equivalent resultant force is zero and the resultant couple moment is M clockwise.

Units Used:

kN = 103 N

Given:

w1 = 4kN/m w2 = 2.5 kN/m

M = 8 kN ∙ m c = 9 m

Units Used:

kN = 103 N

Given:

w1 = 4kN/m w2 = 2.5 kN/m

M = 8 kN ∙ m c = 9 m

Replace the distributed loading by an equivalent resultant force and specify its location, measured from point A.

Units Used:

kN = 103N

Given:

w1 = 800N/m

w2 = 200N/m

a = 2 m

b = 3 m

Units Used:

kN = 103N

Given:

w1 = 800N/m

w2 = 200N/m

a = 2 m

b = 3 m

The distribution of soil loading on the bottom of a building slab is shown. Replace this loading by an equivalent resultant force and specify its location, measured from point O.

Units Used:

kip = 103lb

Given:

w1 = 50 lb/ft

w2 = 300 lb/ft

w3 = 100 lb/ft

a = 12 ft

b = 9 ft

Units Used:

kip = 103lb

Given:

w1 = 50 lb/ft

w2 = 300 lb/ft

w3 = 100 lb/ft

a = 12 ft

b = 9 ft

The beam is subjected to the distributed loading. Determine the length b of the uniform load and its position a on the beam such that the resultant force and couple moment acting on the beam are zero.

Given:

w1 = 40lb/ft c = 10ft

w2 = 60lb/ft = 6 ft

Given:

w1 = 40lb/ft c = 10ft

w2 = 60lb/ft = 6 ft

Units Used:

kip = 103 lb

Given:

w1 = 800lb/ft

w2 = 500lb/ft

a = 12 ft

b = 9 ft

Replace the distributed loading by an equivalent resultant force and specify where its line of action intersects member AB, measured from A.

Given:

w1 = 200N/m

w2 = 100N/m

w3 = 200N/m

a = 5 m

b = 6 m

Given:

w1 = 200N/m

w2 = 100N/m

w3 = 200N/m

a = 5 m

b = 6 m

Replace the distributed loading by an equivalent resultant force and specify where its line of action intersects member BC, measured from C.

Units Used:

kN = 103N

Given:

w1 = 200N/m

w2 = 100N/m

w3 = 200N/m

a = 5 m

b = 6 m

Units Used:

kN = 103N

Given:

w1 = 200N/m

w2 = 100N/m

w3 = 200N/m

a = 5 m

b = 6 m

Replace the loading by an equivalent resultant force and couple moment acting at point O.

Units Used:

kN = 103 N

Given:

w1 = 7.5kN/m

w2 = 20kN/m

a = 3 m

b = 3 m

c = 4.5 m

Units Used:

kN = 103 N

Given:

w1 = 7.5kN/m

w2 = 20kN/m

a = 3 m

b = 3 m

c = 4.5 m

Determine the equivalent resultant force and couple moment at point O.

Units Used:

kN = 103 N

Given:

a = 3 m

wO = 3kN/m

w(x) = wO(x/a)2

Units Used:

kN = 103 N

Given:

a = 3 m

wO = 3kN/m

w(x) = wO(x/a)2

Wind has blown sand over a platform such that the intensity of the load can be approximated by the function w = w0(x/d) 3, simplify this distributed loading to an equivalent resultant force and specify the magnitude and location of the force, measured from A.

Units Used:

kN = 103N

Given:

w0 = 500N/m

d = 10 m

w(x) = wo(x/d)3

Units Used:

kN = 103N

Given:

w0 = 500N/m

d = 10 m

w(x) = wo(x/d)3

Determine the equivalent resultant force and its location, measured from point O.

Determine the equivalent resultant force acting on the bottom of the wing due to air pressure and specify where it acts, measured from point A.

Given:

a = 3 ft

k = 86lb/ft3

w(x) = kx2

Given:

a = 3 ft

k = 86lb/ft3

w(x) = kx2

Currently eighty-five percent of all neck injuries are caused by rear-ends car collisions. To alleviate this problem, an automobile seat restraint has been developed that provides additional pressure contact with the cranium. During dynamic tests the distribution of load on the cranium has been plotted and shown to be parabolic. Determine the equivalent resultant force and its location, measured from point A.

Given:

a = 0.5 ft

w0 = 12lb/ft

k = 24lb/ft3

w(x) = w0 + kx2

Given:

a = 0.5 ft

w0 = 12lb/ft

k = 24lb/ft3

w(x) = w0 + kx2

Determine the equivalent resultant force of the distributed loading and its location, measured from point A. Evaluate the integrals using Simpson's rule.

Units Used:

kN = 103 N

Given:

c1 = 5

c2 = 16

a = 3

b = 1

Units Used:

kN = 103 N

Given:

c1 = 5

c2 = 16

a = 3

b = 1

Determine the coordinate direction angles of F, which is applied to the end A of the pipe assembly, so that the moment of F about O is zero.

Given:

F = 20 lb

a = 8 in

b = 6 in

c = 6 in

d = 10 in

Given:

F = 20 lb

a = 8 in

b = 6 in

c = 6 in

d = 10 in

Determine the moment of the force F about point O. The force has coordinate direction angles α, β, γ. Express the result as a Cartesian vector.

Given:

F = 20 lb a = 8 in

α = 60 deg b = 6 in

β = 120 deg c = 6 in

γ = 45 deg d = 10 in

Given:

F = 20 lb a = 8 in

α = 60 deg b = 6 in

β = 120 deg c = 6 in

γ = 45 deg d = 10 in

Replace the force at A by an equivalent resultant force and couple moment at point P. Express the results in Cartesian vector form.

Units Used:

kN = 103 N

Given:

a = 4 m

b = 6 m

c = 8 m

d = 4 m

F = (-300 200 -500) N

Units Used:

kN = 103 N

Given:

a = 4 m

b = 6 m

c = 8 m

d = 4 m

F = (-300 200 -500) N

Determine the moment of the force FC about the door hinge at A. Express the result as a Cartesian vector.

Given:

F = 250 N

b = 1 m

c = 2.5 m

d = 1.5 m

e = 0.5 m

θ = 30 deg

Given:

F = 250 N

b = 1 m

c = 2.5 m

d = 1.5 m

e = 0.5 m

θ = 30 deg

Determine the magnitude of the moment of the force FC about the hinged axis aa of the door.

Given:

F = 250 N

b = 1 m

c = 2.5 m

d = 1.5 m

e = 0.5 m

θ = 30 deg

Given:

F = 250 N

b = 1 m

c = 2.5 m

d = 1.5 m

e = 0.5 m

θ = 30 deg

A force F1 acts vertically downward on the Z-bracket. Determine the moment of this force about the bolt axis (z axis), which is directed at angle θ from the vertical.

Given:

F1 = 80 N

a = 100 mm

b = 300 mm

c = 200 mm

θ = 15 deg

Given:

F1 = 80 N

a = 100 mm

b = 300 mm

c = 200 mm

θ = 15 deg

Replace the force F having acting at point A by an equivalent force and couple moment at point C.

Units Used: kip = 103 lb

Given:

F = 50 lb

a = 10 ft

b = 20 ft

c = 15 ft

d = 10 ft

e = 30 ft

Units Used: kip = 103 lb

Given:

F = 50 lb

a = 10 ft

b = 20 ft

c = 15 ft

d = 10 ft

e = 30 ft

The horizontal force F acts on the handle of the wrench. What is the magnitude of the moment of this force about the z axis?

Given:

F = 30 N

a = 50 mm

b = 200 mm

c = 10 mm

θ = 45 deg

Given:

F = 30 N

a = 50 mm

b = 200 mm

c = 10 mm

θ = 45 deg

The horizontal force F acts on the handle of the wrench. Determine the moment of this force about point O. Specify the coordinate direction angles α, β, γ of the moment axis.

Given:

F = 30 N c = 10 mm

a = 50 mm θ = 45 deg

b = 200 mm

Draw the free-body diagram of the sphere of weight W resting between the smooth inclined planes. Explain the significance of each force on the diagram.

Given:

W = 10 lb

θ1 = 105 deg

θ2 = 45 deg

Draw the free-body diagram of the hand punch, which is pinned at A and bears down on the smooth surface at B.

Given:

F = 8 lb

a = 1.5 ft

b = 0.2 ft

c = 2 ft

Given:

F = 8 lb

a = 1.5 ft

b = 0.2 ft

c = 2 ft

If the resultant couple moment of the three couples acting on the triangular block is to be zero, determine the magnitudes of forces F and P.

Given:

F1 = 10 lb

a = 3 in

b = 4 in

c = 6 in

d = 3 in

θ = 30 deg

Given:

F1 = 10 lb

a = 3 in

b = 4 in

c = 6 in

d = 3 in

θ = 30 deg

Draw the free-body diagram of the beam supported at A by a fixed support and at B by a roller.

Explain the significance of each force on the diagram.

Given:

w = 40lb/ft

a = 3 ft

b = 4 ft

θ = 30 deg

Explain the significance of each force on the diagram.

Given:

w = 40lb/ft

a = 3 ft

b = 4 ft

θ = 30 deg

Draw the free-body diagram of the jib crane AB, which is pin-connected at A and supported by member (link) BC.

Units Used:

kN = 103 N

Given:

F = 8kN

a = 3 m

b = 4 m

c = 0.4 m

d = 3

e = 4

Units Used:

kN = 103 N

Given:

F = 8kN

a = 3 m

b = 4 m

c = 0.4 m

d = 3

e = 4

Draw the free-body diagram of the C-bracket supported at A, B, and C by rollers. Explain the significance of each force on the diagram.

Given:

a = 3 ft

b = 4 ft

θ1 = 30 deg

θ2 = 20 deg

F = 200 lb

Draw the free-body diagram of the smooth rod of mass M which rests inside the glass. Explain the significance of each force on the diagram.

Given:

M = 20 gm

a = 75 mm

b = 200 mm

θ = 40 deg

Given:

M = 20 gm

a = 75 mm

b = 200 mm

θ = 40 deg

Draw the free-body diagram of the â€œspanner wrenchâ€ subjected to the force F. The support at

A can be considered a pin, and the surface of contact at B is smooth. Explain the significance of each force on the diagram.

Given:

F = 20 lb

a = 1 in

b = 6 in

A can be considered a pin, and the surface of contact at B is smooth. Explain the significance of each force on the diagram.

Given:

F = 20 lb

a = 1 in

b = 6 in

Draw the free-body diagram of the automobile, which is being towed at constant velocity up the incline using the cable at C. The automobile has a mass M and center of mass at G. The tires are free to roll. Explain the significance of each force on the diagram.

Units Used:

Mg = 103 kg

Given:

M = 5 Mg d = 1.50 m

a = 0.3 m e = 0.6 m

b = 0.75 m θ1 = 20 deg

c = 1 m θ2 = 30 deg

g 9.81m/s2

Units Used:

Mg = 103 kg

Given:

M = 5 Mg d = 1.50 m

a = 0.3 m e = 0.6 m

b = 0.75 m θ1 = 20 deg

c = 1 m θ2 = 30 deg

g 9.81m/s2

Draw the free-body diagram of the uniform bar, which has mass M and center of mass at G. The supports A, B, and C are smooth.

Given:

M = 100 kg

a = 1.75 m

b = 1.25 m

c = 0.5 m

d = 0.2 m

g = 9.81m/s2

Given:

M = 100 kg

a = 1.75 m

b = 1.25 m

c = 0.5 m

d = 0.2 m

g = 9.81m/s2

The sphere of weight W rests between the smooth inclined planes. Determine the reactions at the supports.

Given:

W = 10 lb

θ1 = 105 deg

θ2 = 45 deg

Given:

W = 10 lb

θ1 = 105 deg

θ2 = 45 deg

Draw the free-body diagram of the beam, which is pin-connected at A and rocker-supported at B.

Given:

F = 500 N

M = 800 N ∙ m

a = 8 m

b = 4 m

c = 5 m

Given:

F = 500 N

M = 800 N ∙ m

a = 8 m

b = 4 m

c = 5 m

Determine the magnitude of the resultant force acting at pin A of the hand punch.

Given:

F = 8 lb

a = 1.5 ft

b = 0.2 ft

c = 2 ft

Given:

F = 8 lb

a = 1.5 ft

b = 0.2 ft

c = 2 ft

The C-bracket is supported at A, B, and C by rollers. Determine the reactions at the supports.

Given:

a = 3 ft

b = 4 ft

θ1 = 30 deg

θ2 = 20 deg

F = 200 lb

Given:

a = 3 ft

b = 4 ft

θ1 = 30 deg

θ2 = 20 deg

F = 200 lb

The smooth rod of mass M rests inside the glass. Determine the reactions on the rod.

Given:

M = 20 gm

a = 75 mm

b = 200 mm

θ = 40 deg

g = 9.81m/s2

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