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University Physics | 13th Edition | ISBN: 9780321675460 | Authors: Hugh D. Young, Roger A. Freedman
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One end of a uniform meter stick is placed against a

Chapter 5, Problem 70P

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QUESTION:

One end of a uniform meter stick is placed against a vertical wall (?Fig. P11.66?). 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 0.40. (a) What is the maximum value the angle ? can have if the stick is to remain in equilibrium? (b) Let the angle ? be 15o. 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 ? = 15o, 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?

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QUESTION:

One end of a uniform meter stick is placed against a vertical wall (?Fig. P11.66?). 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 0.40. (a) What is the maximum value the angle ? can have if the stick is to remain in equilibrium? (b) Let the angle ? be 15o. 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 ? = 15o, 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?

ANSWER:

Solution 70P Step 1 of 3: Taking torques about the right end of the stick, the friction force is half the weight of the stick, f = w/2. Taking torques about the point where the cord is attached to the wall (the tension in the cord and the friction force exert no torque about this point), and noting that the moment arm of w 1 0 the normal force is l tan , n tan = . Then (2ln) = tan < 0.40, so < tan (0.40) = 22

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