One end of a uniform meter stick is placed against a vertical wall . The other e

Question

One end of a uniform meter stick is placed against a vertical wall . 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 0.40.What is the maximum value the angle heta can have if the stick is to remain in equilibrium?

Let the angle between the cord and the stick is theta = 15degrees 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? When theta = 15 degrees, how large must the coefficient of static friction be so that the block can be attached 10{cm} from the left end of the stick without causing it to slip?

Note: One can only take the torque once, i.e., one can only take the torque about only one point.

Explanation / Answer

1) At the wall there is a friction force, F, up, and a normal force, N, to the left (in my mind it was left). The cord applies a tension force, T, at an angle theta. Rotational equilibrium around the axis centered where the cord is attached requires that mg = 2F since the friction force is exerted at a moment arm twice that of gravity. Equilibrium of vertical components of forces gives F + T sin theta = mg. Horizontal components gives T cos theta = N . At maximum angle the friction will be as large as possible given the normal force N, so F = mu N. Substituting these equations appropriately gives tan theta = mu, so theta = arctan( mu ) = 21.8 degrees.

2) The first two equations are modified to mg + 2 mg(1 - x) = 2F and F + T sin theta = 2 mg, with the other two remaining the same. Substitution leads to - 2x + (3 - x)(tan theta)/mu = 1, which has a solution of .378 m or 37.8 cm.

For 3) you can use the same equations as for 2) as a starting point. Note that in all cases mg will cancel at the end of the day.

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