Vector Mechanics for Engineers: Statics
Vector Mechanics for Engineers: Statics
12th Edition
ISBN: 9781259977268
Author: Ferdinand P. Beer, E. Russell Johnston Jr., David Mazurek
Publisher: McGraw-Hill Education
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Chapter 4.1, Problem 4.23P

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

Chapter 4.1, Problem 4.23P, 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 4.23P, 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 P4.23(a).

Answer to Problem 4.23P

The reaction at A is 44.7lb and is directed along 26.6° above negative x axis and reaction at B is 30.0lb 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 Ax and Ay are the component of the reaction A.

The free body diagram is sketched below as figure 1.

Vector Mechanics for Engineers: Statics, Chapter 4.1, Problem 4.23P , additional homework tip  1

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

Write the expression for the moment at A

MA=F×D (I)

Here, MA is the net moment at A, F is the force, and D is the perpendicular distance between the A 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 A is due to the reaction B and forces 50lb and 40lb.

Thus, the complete expression of MA is

MA=B(20in.)(50lb)(4in.)(40lb)(10in.) (II)

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

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

MA=B(20in.)(50lb)(4in.)(40lb)(10in.)=0 (III)

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

Fx=Ax+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=Ax+40lb=0 (V)

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

Fy=Ay+B50lb=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=Ay+B50lb=0 (VII)

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

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

A=Ax2+Ay2 (VIII)

Here, A is the magnitude of net reaction at A.

Therefore, write the expression for the tanα.

tanα=AyAx (IX)

Calculation:

Rearrange equation (III) to get B.

B(20in.)(50lb)(4in.)(40lb)(10in.)=0B=30lb

From figure 1, the reaction B acts in the perpendicular direction.

Rearrange equation (V) to get Ax .

Ax+40lb=0Ax=40lb

The negative sign indicates that the Ax acts in opposite direction to that assumed initially. Therefore Ax acts in the negative x direction.

Rearrange equation (VII) to get Ay.

Ay=50lbB

Substitute 30.0lb for B in above equation to get Ay.

Ay=50lb30.0lb=20lb

Substitute 40lb for Ax and 20lb for Ay in the equation (VIII) to get A.

A=(40lb)2+(20lb)2=44.7lb

Substitute 40lb for Ax and 20lb for Ay in the equation (IX) to get α.

tanα=20lb40lbα=26.6°

Therefore, the reaction at A is 44.7lb and is directed along 26.6° above negative x axis and reaction at B is 30.0lb 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 4.23P

The reaction at A is 30.2lb and is directed 41.4° above negative x axis and reaction at B is 34.6lb and is directed 60.0° 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 Ax and Ay are the component of the reaction A.

The free body diagram is sketched below as figure 2.

Vector Mechanics for Engineers: Statics, Chapter 4.1, Problem 4.23P , additional homework tip  2

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

Write the expression for the moment at A

MA=F×D (I)

Here, MA is the net moment at A, F is the force, and D is the perpendicular distance between the A 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 A is due to the y component of the reaction B and forces 50lb and 40lb.

Thus, the complete expression of MA is

MA=(Bcos30°)(20in.)(50lb)(4in.)(40lb)(10in.) (II)

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

From figure 2 , u component of reaction B is Bcos30°.

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

MA=(Bcos30°)(20in.)(50lb)(4in.)(40lb)(10in.)=0 (III)

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

Fx=AxBsin30°+40lb (IV)

Here, Fx is the sum of all force along the x direction, Bsin30° is the component of the reaction B along x direction.

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

Fx=AxBsin30°+40lb=0 (V)

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

Fy=Ay+Bcos30°50lb (VI)

Here, Fy is the sum of all force along the y direction, Bcos30° is the component of the reaction B along y direction.

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

Fy=Ay+Bcos30°50lb=0 (VII)

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

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

A=Ax2+Ay2 (VIII)

Here, A is the magnitude of net reaction at A.

Therefore, write the expression for the tanα.

tanα=AyAx (IX)

Calculation:

Rearrange equation (III) to get B.

(Bcos30°)(20in.)(50lb)(4in.)(40lb)(10in.)=0B=34.64lb

From figure 2, the reaction B is 60° above negative x axis.

Rearrange equation (V) to get Ax .

Ax=Bsin30°40lb

Substitute 34.64lb for B in the above equation to get Ax.

Ax=(34.64lb)sin30°40lb=22.68lb

Ax acts in the x direction.

Rearrange equation (VII) to get Ay.

Ay=50lbBcos30°

Substitute 34.64lb for B in the above equation to get Ay.

Ay=50lb(34.64)cos30°=20lb

Substitute 22.68lb for Ax and 20lb for Ay in the equation (VIII) to get A.

A=(22.68lb)2+(20lb)2=30.23lb

Substitute 22.68lb for Ax and 20lb for Ay in the equation (IX) to get α.

tanα=20lb22.68lbα=41.4°

Therefore, the reaction at A is 30.2lb and is directed 41.4° above negative x axis and reaction at B is 34.6lb and is directed 60.0° above negative x axis

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Chapter 4 Solutions

Vector Mechanics for Engineers: Statics

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