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

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 ? with the stick. The coefficient of static friction between the end of the meter stick and the wall is 0.356.

I already did part A.

A.) What is the maximum value the angle ? can have if the stick is to remain in equilibrium? 19.6degrees

B.) Let the angle between the cord and the stick is ? = 14.8?. A block of the same weight as the meter stick is suspended from the stick, as shown, at a distance x from the wall. What is the minimum value of x for which the stick will remain in equilibrium?

C.) When ? = 14.8?, how large must the coefficient of static friction be so that the block can be attached 13.8cm from the left end of the stick without causing it to slip?

Explanation / Answer

B)

Tsin(theta)*1=mg*x+mg*1/2 {torque balance about wall hinge}

also

Tcos(theta)=N {N is towards right from wall on stick}

f=0.356N {f is frictional force acting upwards}

Tsin(theta)+f=2mg

theta=14.8 degrees

On solving we get

mg(x+0.5)(sin(theta)+kcos(theta))=2mgsin(theta) {where k=0.356}

So

x=0.352m

C)

Now x=0.138

So

again solvin above equations we get,

k=coeff. of friction=0.564

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