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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Textbook Question
Chapter 3.3, Problem 3.79P

Solve part a of Prob. 3.78, assuming that two 15-N vertical forces have been added, one acting upward at A and the other downward at C.

Chapter 3.3, Problem 3.79P, Solve part a of Prob. 3.78, assuming that two 15-N vertical forces have been added, one acting

(a)

Expert Solution
Check Mark
To determine

The net couple acting on the rectangular co-ordinates.

Answer to Problem 3.79P

The net couple acting on the rectangular co-ordinates is 8.78Nm.

Explanation of Solution

Write the equation of the couple acting at the reactangular side B and C (Refer fig P3.78).

M1=Fd

Here, moment of couple at the point B and C is M1, the force acting couple at the point B and C is F, and the perpendicular distance from the point B and C to the line of action of force is d.

Substitute 25N for F and 0.75m for d.

M1=(25N)(0.75m)=18.75Nm

Write the equation of the couple acting at the reactangular side D and E (Refer fig P3.78).

M2=Fd

Here, moment of couple at the point D and E is M2, the force acting couple at the point D and E is F, and the perpendicular distance from the point D and E to the line of action of force is d.

Substitute 20N for F and 0.4m for d.

M2=(20N)(0.4m)=8.0Nm

Write the equation of the couple vector acting at the reactangular co-ordinates at point B and C (Refer fig P3.78).

M1=(M1sinθ)i^+(M1cosθ)j^

Here, the projection angle is θ represented in x and y axis.

Substitute 18.75Nm for M1 and 40° for θ.

M1=[(18.75Nm)sin40°]i^+[(18.75Nm)cos40°]j^=(12.0523Nm)i^+(14.3633Nm)j^ (I)

Write the equation of the couple vector acting at the reactangular co-ordinates at point D and E (Refer fig P3.78).

M2=(M2cosθ)j^

Substitute 8.0Nm for M2 and 0° for θ.

M2=[(8.0Nm)cos0°]j^=(8.0Nm)j^ (II)

Sketch the moment of additional couple is added to the moment about A of the force applied at C is shown in the Figure 1.

Vector Mechanics for Engineers: Statics, Chapter 3.3, Problem 3.79P

Write the equation of the additional couple vector acting at the point A of the force applied at C.

M3=Fy×(dxdz)

Here, the additional couple vector formed by the force at the point A and C is M3, force at the point A and C is Fy, perpendicular distance at the point A to the line of action of force is dx, and perpendicular distance at the point C to the line of action of force is dz.

The negative sign of the force shows that the force acting downward direction.

Substitute 15N for Fy, 0.4m for dx, and 0.75m for dz.

M3=(15N)j^×[(0.4m)i^(0.75m)k^]=(11.25Nm)i^(6Nm)k^ (III)

Conclusion:

Write the equation for the net couple acting on the rectangular co-ordinates.

M=M1+M2+M3

Adding the equations (I), (II), and (III) to obtained the net couple.

M=(12.0523Nm)i^+(14.3633Nm)j^(8.0Nm)j^(11.25Nm)i^(6Nm)k^=(0.80Nm)i^+(6.36Nm)j^(6Nm)k^

Write the equation for the magnitude of net couple.

|M|=(0.80Nm)2+(6.36Nm)2+(6Nm)2M=8.780Nm8.78Nm

Therefore, the net couple acting on the rectangular co-ordinates is 8.78Nm.

(b)

Expert Solution
Check Mark
To determine

The direction of the two forces acting on the point B and C that can be used to form couple.

Answer to Problem 3.79P

The direction of the two forces acting on the point B and C is

θx=84.8°, θy=43.6°, θz=133.1°.

Explanation of Solution

Write the equation for direction of the moment of couple formed by the forces at point B and C along x axis.

cosθx=MxM

Here, the angle formed by the moment of couple due to the force located at the point B and C along x axis is θx and moment of couple formed along x axis is Mx.

Rewrite the equation for θx.

θx=cos1(MxM) (IV)

Write the equation for direction of the moment of couple formed by the forces at point B and C along y axis.

cosθy=MyM

Here, the angle formed by the moment of couple due to the force located at the point B and C along y axis is θy and moment of couple formed along y axis is My.

Rewrite the equation for θy.

θy=cos1(MyM) (V)

Write the equation for direction of the moment of couple formed by the forces at point B and C along z axis.

cosθz=MzM

Here, the angle formed by the moment of couple due to the force located at the point B and C along z axis is θz and moment of couple formed along z axis is Mz.

Rewrite the equation for θz.

θz=cos1(MzM) (VI)

Conclusion:

Substitute 0.80Nm for Mx and 8.78Nm for M in equation (IV).

θx=cos1(0.80Nm8.78Nm)=84.8°

Substitute 6.36Nm for My and 8.78Nm for M in equation (V).

θy=cos1(6.36Nm8.78Nm)=43.6°

Substitute 6Nm for Mz and 8.78Nm for M in equation (VI).

θz=cos1(6Nm8.78Nm)=133.1°

Therefore, the direction of the two forces acting on the point B and C is

θx=84.8°, θy=43.6°, θz=133.1°.

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Chapter 3 Solutions

Vector Mechanics for Engineers: Statics

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