VECTOR MECHANICS FOR ENGINEERS W/CON >B
VECTOR MECHANICS FOR ENGINEERS W/CON >B
12th Edition
ISBN: 9781260804638
Author: BEER
Publisher: MCG
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Chapter 4, Problem 4.142RP

(a)

To determine

The reaction at each of the two front wheels A.

(a)

Expert Solution
Check Mark

Answer to Problem 4.142RP

The reaction at each of the two front wheels A is 1761 lb .

Explanation of Solution

The free-body diagram is shown in figure 1.

VECTOR MECHANICS FOR ENGINEERS W/CON >B, Chapter 4, Problem 4.142RP

The weight of the truck is 3200 lb and it is lifting 1700lb crate. The distance at which force acting from the rear wheels are marked in figure 1. Let A be the reaction at one of the front wheel hence, the total reaction across two wheels is 2A.

The reaction at A is acting in the upward direction. The wheel is rotating and there is a torque, which is the product of force acting and perpendicular distance. The moment about B is responsible for the reaction at the front wheel.

Write the expression for the moment at B

MB=F×D (I)

Here, MB is the moment at B, F is the force, and D is the perpendicular distance between the B and the point where force is experienced.

Under equilibrium, the moment about B will balances and sum of all forces will be zero.

ΣMB=(W1×d1)+(W2×d2)(2A×d3) =0 (II)

Here, W1 is the magnitude of the weight of the crate, W2 is the magnitude of the weight of the truck, d1 is the distance from crate to rear wheels, d2 is the distance from center of gravity to rear wheels, d3 is the distance between front and rear wheels, and A is the magnitude of the reaction at a front wheel.

Calculation:

Substitute 1700lb for W1, 52in. for d1, 3200lb for W2 , 110in. for d2, and 36in. for d3 in equation (II) and rearrange to obtain A

ΣMB=(1700 lb)(52 in)+(3200 lb)(12 in)2A(36 in)=0A=(1700lb×52in.)+(3200lb×12in.)(36in)×2=+1761lbA=1761 lb

Therefore, the reaction on each front wheel is A=+1761lb_.

(b)

To determine

The reaction at each of the two rear wheels.

(b)

Expert Solution
Check Mark

Answer to Problem 4.142RP

The reaction at each of the rear wheels is +689lb_.

Explanation of Solution

Refer to figure 1.

Let B be the reaction at one of the rear wheel hence, the total reaction across two wheels is 2B.

The reaction at B is acting in the upward direction. Under equilibrium, the net force in the y direction will be equal to zero.

+ΣFy=0 (III)

Here, ΣFy is the net force in the y direction.

Write the expression for the net force.

+ΣFy=W1W2+2A+2B

Here, W1 is the magnitude of the weight of the crate, W2 is the magnitude of weight of the truck, A is the magnitude of reaction at the front wheel and B is the magnitude of reaction at the rear wheels.

Put the above equation in equation (III).

W1W2+2A+2B=0 (IV)

Calculation:

Substitute 1700lb for W1, 3200lb for W2 and 1761lb for A in the equation (IV) and rearrange to obtain B .

(1700lb)(3200lb)+2(1761lb)+2B=0B=1378 lb2=689 lbB=689 lb

Therefore, the reaction on each rear wheel is +689lb_.

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Chapter 4 Solutions

VECTOR MECHANICS FOR ENGINEERS W/CON >B

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