Chapter 8, Problem 36P

### College Physics

11th Edition
Raymond A. Serway + 1 other
ISBN: 9781305952300

Chapter
Section

### College Physics

11th Edition
Raymond A. Serway + 1 other
ISBN: 9781305952300
Textbook Problem

# Q One end of a uniform 4.0-m-long rod of weight w is supported by a cable at an angle of θ = 37° with the rod. The other end rests against a wall, where it is held by friction. (See Fig. P8.36.) The coefficient of static friction between the wall and the rod is μs = 0.50. Determine the minimum distance x from point A at which an additional weight w (the same as the weight of rod) can be hung without causing the rod to slip at point A.Figure P8.36

To determine

The minimum distance from a point A at which an additional weight w can be hung without causing the rod to slip at point A.

Explanation

Given info: The coefficient of static friction is 0.50 ,angle of the cable with the rod is 37Â° and the force of friction is 0.40T .

The frictional force at point A is f=(fs)max=Î¼sn but the sum of horizontal forces on the rod at equilibrium condition is âˆ‘Fx=0=nâˆ’TcosÎ¸. By using these two expressions, the force of friction is calculated. At equilibrium condition, the vertical components of forces on the rod is defined as âˆ‘Fy=0=f+TsinÎ¸âˆ’2w. The sum of torques about the point A is âˆ‘Ï„=0=âˆ’xminwâˆ’(l/2)w+l(TsinÎ¸)=0.

The formula for the force of friction is,

f=Î¼sTcosÎ¸

• Î¼s is coefficient of static friction.
• T is force of tension in the cable.
• Î¸ is the angle of the cable with the rod.

Substitute 0.50 for Î¼s and 37Â° for Î¸ to find f.

f=(0.50)(cos37Â°)T=0.40T

Thus, the force of friction the surfaces of the wall and the end of the rod is 0.40T

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