Physics for Scientists and Engineers
Physics for Scientists and Engineers
10th Edition
ISBN: 9781337553278
Author: Raymond A. Serway, John W. Jewett
Publisher: Cengage Learning
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Textbook Question
Chapter 12, Problem 35AP

A uniform beam of mass m is inclined at an angle θ to the horizontal. Its upper end (point P) produces a 90° bend in a very rough rope tied to a wall, and its lower end rests on a rough floor (Fig. P12.35). Let μs represent the coefficient of static friction between beam and floor. Assume μs is less than the cotangent of θ. (a) Find an expression for the maximum mass M that can be suspended from the top before the beam slips. Determine (b) the magnitude of the reaction force at the floor and (c) the magnitude of the force exerted by the beam on the rope at P in terms of m, M, and μs.

Figure P12.35

Chapter 12, Problem 35AP, A uniform beam of mass m is inclined at an angle  to the horizontal. Its upper end (point P)

 (a)

Expert Solution
Check Mark
To determine

The expression for the maximum mass that can be suspended from the top surface before the beam starts slip.

Answer to Problem 35AP

The expression for the maximum mass that can be suspended from the top surface before the beam starts slip is 12m(2μssinθcosθcosθμssinθ).

Explanation of Solution

The mass of the beam is m, inclination angle of the beam is θ, the coefficient of static friction between the beam and the floor is μs, and the μs is less than cotangent of θ.

The following figure shows the force diagram of the beam.

Physics for Scientists and Engineers, Chapter 12, Problem 35AP

Figure-(I)

Formula to calculate the frictional force acting on the base of the beam is,

    f=μsn

Here, f is the frictional force acting on the base of the beam, μs is the coefficient of static friction and n is the normal force acting at point O.

Formula to calculate the net torque about the point O is,

    mg(L2cosθ)+Mg(Lcosθ)T(Lsinθ)=0

Here, m is the mass of the beam, g is the acceleration due to gravity, L is the length of the beam, θ is the inclination angle of the beam with horizontal, M is the mass hanged at the upper end of the beam and T is the tension on the rope.

Rearrange the above equation to find T.

    mg(L2cosθ)+Mg(Lcosθ)T(Lsinθ)=0T(Lsinθ)=mg(L2cosθ)+Mg(Lcosθ)T=mgcosθ2sinθ+Mgcosθsinθ

Formula to calculate the net vertical forces is,

    Mg+mgn=0

Rearrange the above equation to find n.

    Mg+mgn=0n=Mg+mg

Formula to calculate the net horizontal force is,

    T=f

Substitute mgcosθ2sinθ+Mgcosθsinθ for T, μsn for f and Mg+mg for n in the above equation to find M.

    (mgcosθ2sinθ+Mgcosθsinθ)=μs(Mg+mg)mcosθ2sinθ+Mcosθsinθ=μsM+mμsM(cosθsinθμs)=m(μscosθ2sinθ)M=m(μscosθ2sinθ)(cosθsinθμs)

Further simplify the above equation to find M.

    M=m(μscosθ2sinθ)(cosθsinθμs)M=m2(2μssinθcosθ)(cosθμssinθ)

Conclusion:

Therefore, the expression for the maximum mass that can be suspended from the top surface before the beam starts slip is 12m(2μssinθcosθcosθμssinθ).

(b)

Expert Solution
Check Mark
To determine

The magnitude of the reaction force at the floor.

Answer to Problem 35AP

The magnitude of the reaction force at the floor is (m+M)g1+μs2.

Explanation of Solution

The mass of the beam is m, inclination angle of the beam is θ, the coefficient of static friction between the beam and the floor is μs, and the μs is less than cotangent of θ.

Formula to calculate the net reaction force acting in the floor is,

    R=n2+f2

Here, R is the reaction force at the floor.

Substitute Mg+mg for n and μsn for f in the above equation to find R.

    R=(Mg+mg)2+(μs(Mg+mg))2=(Mg+mg)2(1+μs2)=(Mg+mg)(1+μs2)=g(m+M)1+μs2

Conclusion:

Therefore, the magnitude of the reaction force at the floor is (m+M)g1+μs2.

(c)

Expert Solution
Check Mark
To determine

The magnitude of the force exerted by the beam in the rope at P.

Answer to Problem 35AP

The magnitude of the force exerted by the beam in the rope at P is gM2+μs2(m+M)2.

Explanation of Solution

The mass of the beam is m, inclination angle of the beam is θ, the coefficient of static friction between the beam and the floor is μs, and the μs is less than cotangent of θ.

Since the coefficient of static friction of the floor is less than cotangent of the angle θ, thus,

    fμsn

Substitute mgcosθ2sinθ+Mgcosθsinθ for f and Mg+mg for n in the above equation to find μs.

    (mgcosθ2sinθ+Mgcosθsinθ)μs(Mg+mg)μs[M+m2M+m]cotθ

Formula to calculate the magnitude of the net force exerted by the beam in the rope at point P is,

    F=T2+(Mg)2

Here, F is the magnitude of the force exerted by the beam in the rope.

Substitute mgcosθ2sinθ+Mgcosθsinθ for T in the above equation to find F.

    F=(mgcosθ2sinθ+Mgcosθsinθ)2+(Mg)2=(m2+M)2g2cot2θ+(Mg)2=g(m2+M)2cot2θ+M2=gM2+(m+M)2[(m2+Mm+M)cotθ]2

Substitute μs for [M+m2M+m]cotθ in the above equation to find F.

    F=gM2+(m+M)2[(m2+Mm+M)cotθ]2=gM2+(m+M)2[μs]2=gM2μs2(m+M)2

Conclusion:

Therefore, the magnitude of the force exerted by the beam in the rope at P is gM2+μs2(m+M)2

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Chapter 12 Solutions

Physics for Scientists and Engineers

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