   Chapter 8, Problem 36P

Chapter
Section
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 37° and the force of friction is 0.40T .

The frictional force at point A is f=(fs)max=μsn but the sum of horizontal forces on the rod at equilibrium condition is Fx=0=nTcosθ. 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 Fy=0=f+Tsinθ2w. The sum of torques about the point A is τ=0=xminw(l/2)w+l(Tsinθ)=0.

The formula for the force of friction is,

f=μsTcosθ

• μs is coefficient of static friction.
• T is force of tension in the cable.
• θ is the angle of the cable with the rod.

Substitute 0.50 for μs and 37° for θ to find f.

f=(0.50)(cos37°)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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