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Fundamentals Of Physics
Found in: Page 346
Fundamentals Of Physics

Fundamentals Of Physics

Book edition 10th Edition
Author(s) David Halliday
Pages 1328 pages
ISBN 9781118230718

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Short Answer

The system in Fig. 12-38 is in equilibrium. A concrete block of mass225 kg hangs from the end of the uniform strut of mass45.0 kg. A cable runs from the ground, over the top of the strut, and down to the block, holding the block in place. For anglesϕ = 30.0° andθ= 45.0° , find (a) the tension T in the cable and the (b) horizontal and (c) vertical components of the force on the strut from the hinge.

a)The tension in the cable is.6.63×103N

b)Horizontal component of the force is5.74×103N

c) Vertical component of the force is5.96×103N

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Step by Step Solution

TABLE OF CONTENTS :
TABLE OF CONTENTS

Step 1: Listing the given quantities

Theconcrete block of mass225 kg

Strut of mass45.0 kg

ϕ = 30.0°θ= 45.0°

Step 2: Understanding the concept of the horizontal and vertical componentof the force

In the free body diagram of the given situation, we notice the acting torques on the body both horizontally and vertically.Here, the acting torque results from the applied force along the length of the cable hanging from the strut and along the weight of the block to balance it at the strut.The vertical force acting on the body about the hinge is due to the tension resulting from the hanging to the strut, also the downward pull due to the weight of the block, and the downward pull due to the weight of the strut that acts at the center of the strut. Then using Newton's laws of motion, we balance the acting torques in an equation separately for both the horizontal and the vertical directions. Then, accordingly, calculate the required values as per the problem

Step 3: Calculation of the tension in the cable

We note that the angle between the cable and the strut is

α=θ-ϕ = 45°-30°=15°

The angle between the strut and any vertical force(like the weights in the problem) is

β= 90°-45°=45°

DenotingM= 225kgandm=45.0kgand ℓas the length of the boom, we compute torques about the hinge and find

T=Mgℓsinβ+mgℓ2sinβℓsinα=Mgsinβ+mgsinβ2sinα

The unknown lengthcancels out and we obtainT= 6.63×103N

The tension in the cable is.6.63×103N

Step 4:  Calculation of thehorizontal component of the force

(b)

Since the cable is at30° from horizontal, then horizontal equilibrium of forces requires that the horizontal hinge force be

Fx=Tcos30°=5.74×103N

Horizontal component of the force is5.74×103N

Step 5: Calculation of thevertical component of the force

(c)

Vertical equilibrium of forces gives the vertical hinge force component

Fy=Mg+mg+Tsin30°=5.96×103N

Vertical component of the force is5.96×103N

Most popular questions for Physics Textbooks

A uniform ladder whose length is 5.0 m and whose weight is 400 N leans against a frictionless vertical wall. The coefficient of static friction between the level ground and the foot of the ladder is 0.46. What is the greatest distance the foot of the ladder can be placed from the base of the wall without the ladder immediately slipping?

Figure 12-59 shows the stress versus strain plot for an aluminum wire that is stretched by a machine pulling in opposite directions at the two ends of the wire. The scale of the stress axis is set by s=7.0, in units of 107 N/m2. The wire has an initial length of 0.800 mand an initial cross-sectional area of 2.00×10−6 m2. How much work does the force from the machine do on the wire to produce a strain of1.00×10−3?

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Question: Figure 12-29 shows a diver of weight 580 N standing at the end of a diving board with a length of L =4.5 m and negligible v mass. The board is fixed to two pedestals (supports) that are separated by distance d = 1 .5 m . Of the forces acting on the board, what are the (a) magnitude and (b) direction (up or down) of the force from the left pedestal and the (c) magnitude and (d) direction (up or down) of the force from the right pedestal? (e) Which pedestal (left or right) is being stretched, and (f) which pedestal is being compressed?

A uniform beam is 5.0 m long and has a mass of 53 kg. In Fig. 12-74, the beam is supported in a horizontal position by a hinge and a cable, with angleθ=60° . In unit-vector notation, what is the force on the beam rom the hinge?

A makeshift swing is constructed by making a loop in one end of a rope and tying the other end to a tree limb. A child is sitting in the loop with the rope hanging vertically when the child’s father pulls on the child with a horizontal force and displaces the child to one side. Just before the child is released from rest, the rope makes an angle of 15° with the vertical and the tension in the rope is 280 N .

(a) How much does the child weigh?

(b) What is the magnitude of the (horizontal) force of the father on the child just before the child is released?

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