Physics for Scientists and Engineers, Technology Update (No access codes included)
Physics for Scientists and Engineers, Technology Update (No access codes included)
9th Edition
ISBN: 9781305116399
Author: Raymond A. Serway, John W. Jewett
Publisher: Cengage Learning
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Chapter 12, Problem 12.66CP

In the What If? section of Example 12.2, let d represent the distance in meters between the person and the hinge at the left end of the beam. (a) Show that the cable tension is given by T = 93.9d + 125, with T in newtons. (b) Show that the direction angle θ of the hinge force is described by

tan θ = ( 32 3 d + 4 1 ) tan 53.0 °

(c) Show that the magnitude of the hinge force is given by

R = 8.82 × 10 3 d 2 9.65 × 10 4 d + 4.96 × 10 5

(d) Describe how the changes in T, θ, and R as d increases differ from one another.

(a)

Expert Solution
Check Mark
To determine

The cable tension is T=93.9d+125.

Answer to Problem 12.66CP

The cable tension is given by, T=93.9d+125.

Explanation of Solution

The weight of the person is 600N, the weight of the beam is 200N, the length of the beam is 8m, cables makes an angle of 53°, distance between the person and hinged support is d, the tension in the rope is T.

The diagram for the given condition is shown below.

Physics for Scientists and Engineers, Technology Update (No access codes included), Chapter 12, Problem 12.66CP

Figure (1)

Apply the rotational equilibrium equation and take the torque about hinge support.

    (Tsinϕ)×lwp×dwb×l2=0        (1)

Here, wp is the weight of the person, wb is the weight of the beam, l is the length of the beam an d ϕ is the cable angle.

Rearrange the equation (1) for T.

    T=wp×d+wb×l2l×sinϕ        (2)

Substitute, 600N for wp, 200N for wb, 8m for l and 53° for ϕ in equation (2) to find the expression for the tension.

    T=600N×d+200N×828×sin53T=93.9d+125

Conculasion:

Therefore, the cable tension will be T=93.9d+125.

(b)

Expert Solution
Check Mark
To determine

The direction angle is tanθ=(323d+41)tan53°

Answer to Problem 12.66CP

The direction angle will be tanθ=(323d+41)tan53°.

Explanation of Solution

Apply the equilibrium condition for the horizontal forces.

    Fx=0RcosθTcosϕ=0Rcosθ=Tcosϕ        (3)

Apply the equilibrium equation for the vertical forces.

    Fy=0Rsinθ+Tsinϕwpwb=0Rsinθ=wp+wbTsinϕ        (4)

Here, R is the hinge reaction and θ is the angle makes by the hinge reaction.

Divide equation (4) by equation (3).

    RsinθRcosθ=wp+wbTsinϕTcosϕtanθ=(wp+wbTsinϕTcosϕ)×tanϕtanϕtanθ=(wp+wbTsinϕTcosϕ)tanϕtanθ=(wp+wbTsinϕ1)tanϕ        (5)

Substitute, 600N for wp, 200N for wb, (93.9d+125)N for T  and 53° for ϕ in equation (5) to find the reaction angle.

    tanθ=(600N+200N(93.9d+125)N×sin53°1)tan53°tanθ=(323d+41)tan53°

Conculasion:

Therefore, The direction angle will be tanθ=(323d+41)tan53°

(c)

Expert Solution
Check Mark
To determine

The magnitude of the hinge force is R=8.82×105d29.65×104d+4.96×105 .

Answer to Problem 12.66CP

The magnitude of the hinge force will be R=8.82×105d29.65×104d+4.96×105.

Explanation of Solution

Square on both side of equation(2) and equation(3) and then add them.

    R2(cos2θ+sin2θ)=T2cos2ϕ+(wp+wbTsinϕ)2R=T2+(wp+wb)22Tsinϕ(wp+wb)        (6)

Substitute, 600N for wp, 200N for wb, (93.9d+125)N for T  and 53° for ϕ in equation (6) to find the reaction.

    R=(93.9d+125)2N+(600N+200N)22(93.9d+125)N×sin53°(600N+200N)=8.82×105d29.65×104d+4.96×105

Conclusion:

Therefore, the magnitude of the hinge force is R=8.82×105d29.65×104d+4.96×105.

(d)

Expert Solution
Check Mark
To determine

The changes in T, θ and R on increasing the distance d.

Answer to Problem 12.66CP

T and R increases on increases the distance d and θ decreases when the distance d increases.

Explanation of Solution

T and R is the directely proportional to the distance d. since when d is increases then T and R is increases. But angle θ is inversely proportional to the distance d so when the d is increases then the magnitude of the angle is decreases.

Conclusion:

Therefore, T and R increases on increases the distance d and θ decreases.

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Chapter 12 Solutions

Physics for Scientists and Engineers, Technology Update (No access codes included)

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