One end of a uniform meter stick is placed against a vertical wall (Figure 1). T

Question

One end of a uniform meter stick is placed against a vertical wall (Figure 1). The other end is held by a lightweight cord that makes an angle theta with the stick. The coefficient of static friction between the end of the meter stick and the wall is 0.37. Let the angle between the cord and the stick is theta = 11 degree. A block of the same weight as the meter stick is suspended from the stick, as shown at (Figure 2), at a distance x from the wall. What is the minimum value of x for which the stick will remain in equilibrium? When theta = 11 degree, how large must the coefficient of static friction be so that the block can be attached 12 cm from the left end of the stick without causing it to slip?

Explanation / Answer

along vertical

T*sintheta + u*T*costheta = 2W

T*sin11 + (0.37*T*cos11) = 2W

T = 3.61*W

net torque about left end = 0

W*(x) + W*L/2 = T*sintheta*L

W*(x) + W*L/2 = 3.61*W*sintheta*L

x + 1/2 = 3.61*sin11*1

x = 18.8 cm

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part(c)

for x = 12 cm = 0.12 m

W*(x) + W*L/2 = T*sintheta*L

(W*0.12) + (W/2) = T*sin11*1

T = 3.25 W

along vertical

T*sintheta + u*T*costheta = 2W

3.25*W*sin11 + (u*3.25*W*cos11) = 2W

(3.25*sin11) + (u*3.25*cos11) = 2

u = 0.432

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