I am confused about why Ff - w + Fr is in the x-direction.I feel like...

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I am confused about why Ff - w + Fr is in the x-direction.I feel like there is something I am missing about torque and forces. Based on the drawing of the forces in the picture, I would assume that these forces were all in the y-direction. Thank you. 12.Static Equilibrium and Elasticity 570 0.52w -d-2.5 m- Figure 12.4 The weight distribution between the axles of a car. Where is the center of gravity located? (credit "car": modification of work by Jane Whitney) Strategy We do not know the weight w of the car. All we know is that when the car rests on a level surface, 0.52w pushes down on the surface at contact points of the front wheels and 0.48w pushes down on the surface at contact points of the rear wheels. Also, the contact points are separated from each other by the distance d = 2.5 m. At these contact points, the car experiences normal reaction forces with magnitudes FF = 0.52w and FR = 0.48w on the front and rear axles, respectively. We also know that the car is an example of a rigid body in equilibrium whose entire weight wacts at its CM. The CM is located somewhere between the points where the normal reaction forces act, somewhere at a distance x from the point where FR acts. Our task is to find x. Thus, we identify three forces acting on the body (the car), and we can draw a free-body diagram for the extended rigid body, as shown in Figure 12.5. F = 0.52w Ff=d-x CM 0.48W FR = 0.48w FR=X Figure 12.5 The free-body diagram for the car clearly indicates force vectors acting on the car and distances to the center of mass (CM). When CM is selected as the pivot point, these distances are lever arms of normal reaction forces. Notice that vector magnitudes and lever arms do not need to be drawn to scale, but all quantities of relevance must be clearly labeled. We are almost ready to write down equilibrium conditions Equation 12.7 through Equation 12.2 for the car, but first we must decide on the reference frame. Suppose we choose the x-axis along the length of the car, the y-axis vertical, and the z-axis perpendicular to this xy-plane. With this choice we only need to write Equation 12.7 and Equation 12.2 because all the y-components are identically zero. Now we need to decide on the location of the pivot point. We can choose any point as the location of the axis of rotation (z-axis). Suppose we place the axis of rotation at CM, as indicated in the free-body diagram for the car. At this point, we are ready to write the equilibrium conditions for the car. Solution Each equilibrium condition contains only three terms because there are N = 3 forces acting on the car. The first equilibrium condition, Equation 12.2, reads +FF-W + FR = 0. 12.11 I am confused about why Ff - w + Fr is in the x-direction.I feel like there is something I am missing about torque and forces. Based on the drawing of the forces in the picture, I would assume that these forces were all in the y-direction. Thank you. 12.Static Equilibrium and Elasticity 570 0.52w -d-2.5 m- Figure 12.4 The weight distribution between the axles of a car. Where is the center of gravity located? (credit "car": modification of work by Jane Whitney) Strategy We do not know the weight w of the car. All we know is that when the car rests on a level surface, 0.52w pushes down on the surface at contact points of the front wheels and 0.48w pushes down on the surface at contact points of the rear wheels. Also, the contact points are separated from each other by the distance d = 2.5 m. At these contact points, the car experiences normal reaction forces with magnitudes FF = 0.52w and FR = 0.48w on the front and rear axles, respectively. We also know that the car is an example of a rigid body in equilibrium whose entire weight wacts at its CM. The CM is located somewhere between the points where the normal reaction forces act, somewhere at a distance x from the point where FR acts. Our task is to find x. Thus, we identify three forces acting on the body (the car), and we can draw a free-body diagram for the extended rigid body, as shown in Figure 12.5. F = 0.52w Ff=d-x CM 0.48W FR = 0.48w FR=X Figure 12.5 The free-body diagram for the car clearly indicates force vectors acting on the car and distances to the center of mass (CM). When CM is selected as the pivot point, these distances are lever arms of normal reaction forces. Notice that vector magnitudes and lever arms do not need to be drawn to scale, but all quantities of relevance must be clearly labeled. We are almost ready to write down equilibrium conditions Equation 12.7 through Equation 12.2 for the car, but first we must decide on the reference frame. Suppose we choose the x-axis along the length of the car, the y-axis vertical, and the z-axis perpendicular to this xy-plane. With this choice we only need to write Equation 12.7 and Equation 12.2 because all the y-components are identically zero. Now we need to decide on the location of the pivot point. We can choose any point as the location of the axis of rotation (z-axis). Suppose we place the axis of rotation at CM, as indicated in the free-body diagram for the car. At this point, we are ready to write the equilibrium conditions for the car. Solution Each equilibrium condition contains only three terms because there are N = 3 forces acting on the car. The first equilibrium condition, Equation 12.2, reads +FF-W + FR = 0. 12.11

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