VECTOR MECHANICS FOR ENGINEERS: STATICS
VECTOR MECHANICS FOR ENGINEERS: STATICS
12th Edition
ISBN: 9781260912814
Author: BEER
Publisher: MCG
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Chapter 4.1, Problem 4.24P

4.23 and 4.24 For each of the plates and loadings shown, determine the reaction at A and B.

Chapter 4.1, Problem 4.24P, 4.23 and 4.24 For each of the plates and loadings shown, determine the reaction at A and B. Fig. , example  1

Fig. P4.23

Chapter 4.1, Problem 4.24P, 4.23 and 4.24 For each of the plates and loadings shown, determine the reaction at A and B. Fig. , example  2

Fig. P4.24

(a)

Expert Solution
Check Mark
To determine

The reaction at A and B of the plate shown in P4.24(a).

Answer to Problem 4.24P

The reaction at A is 20.0lb and reaction at B is 50.0lb and is directed along 36.9° above negative x axis in the upward direction.

Explanation of Solution

Take vectors along positive x and y axis are positive.

Let A is the reaction at the point A , B is the reaction at the point B and Bx and By are the component of the reaction B.

The free body diagram is sketched below as figure 1.

VECTOR MECHANICS FOR ENGINEERS: STATICS, Chapter 4.1, Problem 4.24P , additional homework tip  1

Here, Bx and By are the magnitude of x and y component of reaction B at the point A and B is the magnitude of reaction B.

Write the expression for the moment at B.

MB=F×D (I)

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

Above equation implies that net moment at any point is the sum of product of each force acting on the system and perpendicular distance of the force and the point.

The moment at B is due to the reaction A and forces 50lb and 40lb.

Thus, the complete expression of MB is

MB=A(20in.)+(50lb)(16in.)(40lb)(10in.) (II)

Here, MB is the sum of all anticlockwise moment of force at A.

At equilibrium, the sum of the moment acting at A will be zero

MB=A(20in.)+(50lb)(16in.)(40lb)(10in.)=0 (III)

Write the expression for the net force along the x direction.

Fx=Bx+40lb (IV)

Here, Fx is the sum of all force along the x direction.

At equilibrium, the net force along the x direction will be zero.

Fx=Bx+40lb=0 (V)

Write the expression for the net force along the y direction.

Fy=By+A50lb=0 (VI)

Here, Fy is the sum of all force along the y direction.

At equilibrium, the net force along the y direction will be zero.

Fy=By+A50lb=0 (VII)

Let α be the angle that B makes with x axis.

Write the expression for the magnitude of net reaction at B.

B=Bx2+By2 (VIII)

Here, B is the magnitude of net reaction at point B.

Therefore, write the expression for the tanα.

tanα=ByBx (IX)

Calculation:

Rearrange equation (III) to get B.

A(20in.)+(50lb)(16in.)(40lb)(10in.)=0A=+20lb

From figure 1, the reaction A acts in the upward direction.

Rearrange equation (V) to get Bx .

Bx=40lb

The x component of reaction acts in the x direction.

Rearrange equation (VII) to get By.

By=50lbA

Substitute 20.0lb for A in above equation to get By.

By=50lb20.0lb=30.0lb

Substitute 40lb for Bx and 30.0lb for By in the equation (VIII) to get B.

B=(40lb)2+(30lb)2=50.0lb

Substitute 40lb for Bx and 30.0lb for By in the equation (IX) to get α.

tanα=30.0lb40.0lbα=36.9°

Therefore, the reaction at A is 20.0lb and reaction at B is 50.0lb and is directed along 36.9° above negative x axis in the upward direction.

(b)

Expert Solution
Check Mark
To determine

The reaction at A and B of the plate shown in P4.23(b).

Answer to Problem 4.24P

The reaction at A is 23.1lb and is directed 60.0° above positive x axis and reaction at B is 59.6lb and is directed 30.2° above negative x axis.

Explanation of Solution

Take vectors along positive x and y axis are positive.

Let A is the reaction at the point A , B is the reaction at the point B and Bx and By are the component of the reaction B.

The free body diagram is sketched below as figure 1.

VECTOR MECHANICS FOR ENGINEERS: STATICS, Chapter 4.1, Problem 4.24P , additional homework tip  2

Here, Bx and By are the magnitude of x and y component of reaction B at the point A and B is the magnitude of reaction B.

Write the expression for the moment at B.

MB=F×D (I)

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

Above equation implies that net moment at any point is the sum of product of each force acting on the system and perpendicular distance of the force and the point.

The moment at B is due to the y component of reaction A and forces 50lb and 40lb.

Thus, the complete expression of MB is

MB=(Acos30°)(20in.)+(50lb)(16in.)(40lb)(10in.) (II)

Here, MB is the sum of all anticlockwise moment of force at A.

At equilibrium, the sum of the moment acting at A will be zero

MB=(Acos30°)(20in.)+(50lb)(16in.)(40lb)(10in.)=0 (III)

Write the expression for the net force along the x direction.

Fx=Asin30°+Bx+40lb (IV)

Here, Fx is the sum of all force along the x direction.

At equilibrium, the net force along the x direction will be zero.

Fx=Asin30°+Bx+40lb=0 (V)

Write the expression for the net force along the y direction.

Fy=By+Acos30°50lb (VI)

Here, Fy is the sum of all force along the y direction.

At equilibrium, the net force along the y direction will be zero.

Fy=By+Acos30°50lb=0 (VII)

Let α be the angle that B makes with x axis.

Write the expression for the magnitude of net reaction at B.

B=Bx2+By2 (VIII)

Here, B is the magnitude of net reaction at point B.

Therefore, write the expression for the tanα.

tanα=ByBx (IX)

Calculation:

Rearrange equation (III) to get B.

(Acos30°)(20in.)+(50lb)(16in.)(40lb)(10in.)=0A=23.1lb

From figure 1, the reaction A is directed at 60° above positive x axis

Rearrange equation (V) to get Bx .

Bx=40lbAsin30°

Substitute 23.09lb for A in above equation to get Bx.

Bx=40lb(23.09lb)sin30°=51.55lb

The x component of reaction B acts in the x direction.

Rearrange equation (VII) to get By.

By=50lbAcos30°

Substitute 23.09lb for A in above equation to get By.

By=50lb(23.09lb)cos30°=30lb

Substitute 51.55lb for Bx and 30lb for By in the equation (VIII) to get B.

B=(51.55lb)2+(30lb)2=59.64lb

Substitute 51.55lb for Bx and 30lb for By in the equation (IX) to get α.

tanα=30lb51.55lbα=30.2°

Therefore, the reaction at A is 23.1lb and is directed 60.0° above positive x axis and reaction at B is 59.6lb and is directed 30.2° above negative x axis.

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Chapter 4 Solutions

VECTOR MECHANICS FOR ENGINEERS: STATICS

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