ch12-p037

# ch12-p037 - 37. The free-body diagram on the right shows...

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37. The free-body diagram on the right shows the forces acting on the plank. Since the roller is frictionless the force it exerts is normal to the plank and makes the angle θ with the vertical. Its magnitude is designated F . W is the force of gravity; this force acts at the center of the plank, a distance L /2 from the point where the plank touches the floor. N F is the normal force of the floor and f is the force of friction. The distance from the foot of the plank to the wall is denoted by d . This quantity is not given directly but it can be computed using d = h /tan . The equations of equilibrium are: horizontal force components vertical force components torques () 2 sin 0 cos 0 cos 0. N L N Ff FW F Fd fh W d −= −+ = −− = The point of contact between the plank and the roller was used as the origin for writing the torque equation. When = 70º the plank just begins to slip and f = μ sF N , where μ s is the coefficient of static friction. We want to use the equations of equilibrium to compute F N and f for = 70º, then use s = f / F N to compute the coefficient of friction. The second equation gives

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## This note was uploaded on 05/17/2011 for the course PHY 2049 taught by Professor Any during the Spring '08 term at University of Florida.

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ch12-p037 - 37. The free-body diagram on the right shows...

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