Using the force diagram, apply Newton's second law, and sum the torques to find the unknown...

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Using the force diagram, apply Newton's second law, and sum the torques to find the unknown forces. Since this is a static problem (no motion), both the net force and the net torque are zero. Part 3 of 4 - Analyze (a) The wall is frictionless, but it does exert a horizontal normal force n. For the x and y components of the force, we have the following from Newton's second law. Fx=fo-nw=0 y = - - 485 N = 0 Enter a number. Taking torques about an axis at the foot of the ladder, we have the following. =( +( 485 N)(1 n nw² [(3.84 m)(C Solving this equation for N) (3.84 m) s .84 m) sin 35° m sin 350 we have m cos 35° m N. Next substitute the value for w into the F, equation to find N. The friction force is in the positive x direction toward the wall. Submit Skip (you cannot come back) m) (485 N)] tan 35⁰ Using the force diagram, apply Newton's second law, and sum the torques to find the unknown forces. Since this is a static problem (no motion), both the net force and the net torque are zero. Part 3 of 4 - Analyze (a) The wall is frictionless, but it does exert a horizontal normal force n. For the x and y components of the force, we have the following from Newton's second law. Fx=fo-nw=0 y = - - 485 N = 0 Enter a number. Taking torques about an axis at the foot of the ladder, we have the following. =( +( 485 N)(1 n nw² [(3.84 m)(C Solving this equation for N) (3.84 m) s .84 m) sin 35° m sin 350 we have m cos 35° m N. Next substitute the value for w into the F, equation to find N. The friction force is in the positive x direction toward the wall. Submit Skip (you cannot come back) m) (485 N)] tan 35⁰

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Answer Applying Newtons second law to the x and y components of the force In the xdirection the only ... View the full answer