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
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Chapter 3.3, Problem 3.85P

A worker tries to move a rock by applying a 360-N force to a steel bar as shown, (a) Replace that force with an equivalent force-couple system at D. (b) Two workers attempt to move the same rock by applying a vertical force at A and another force at D. Determine these two forces if they are to be equivalent to the single force of Part a.

Chapter 3.3, Problem 3.85P, A worker tries to move a rock by applying a 360-N force to a steel bar as shown, (a) Replace that

(a)

Expert Solution
Check Mark
To determine

An equivalent force-couple system applied at D to replace the force at B.

Answer to Problem 3.85P

Thus the force-couple system at D to replace the force at B is 231.40Ni^275.78Nj^, 230.45Nmk^.

Explanation of Solution

Refer Fig P3.85.

Write the expression of the force at D in component form.

FB=FB(sinθi^cosθj^) (1)

Here FB is the magnitude of the force applied at B, θ is the angle made by the force with the vertical, FB is the force at B.

Write an expression of the vector connecting point B and D.

r=dBD(cosαi^+sinαj^) (2)

Here r is the vector joining point B and D, dBD is the distance between point B and D, α is the angle made by the steel bar with the horizontal.

Write an expression for the couple at D.

M=r×F (3)

Here M is the couple at D, r is the vector joining point B and D.

Conclusion:

Substitute 360N for FB, 40° for θ to determine FB in equation (1).

FB=360N(sin40°i^cos40°j^)=231.40Ni^275.78Nj^

Substitute 0.65m for dBD, 30° for α in equation (2)

r=(0.65m)(cos30°i^+sin30°j^)=0.56292mi^+0.32500mj^

Substitute the value of FB and r for M.

M=|i^j^k^0.56292m0.32500m0231.40N275.78N0|=230.45Nmk^

Thus the force-couple system at D to replace the force at B is 231.40Ni^275.78Nj^, 230.45Nmk^.

(b)

Expert Solution
Check Mark
To determine

Two forces at A and D to replace the force in section (a).

Answer to Problem 3.85P

Thus the vertical force at A is 253.42Nj^, and at D is 232.48N making an angle of 5.5193° with the x-axis.

Explanation of Solution

Refer Fig P3.85.

Write the expression of couple at D.

MD=rAD×FA (1)

Here MD is the couple at D, rAD is the vector joining A and D, FA is the force at A.

Write the expression of the rAD.

rAD=(dAD)(cosαi^+sinαj^) (2)

Here dAD is the distance between A and D, α is the angle between the steel bar with the horizontal.

Write the expression connecting the forces at A, D and B with the force in the section (a).

F=FA+FD (3)

Here F is the force in the section (a), FA is the force at A, FD is the force at D.

Conclusion:

Substitute 1.05m for dAD, 30° for α in equation (2).

rAD=(1.05m)(cos30°i^+sin30°j^)=0.90933mi^+0.52500mj^

Substitute 230.45Nmk^ for MD, 0.90933mi^+0.52500mj^ for rAD in equation (1).

230.45Nmk^=|i^j^k^0.90933m0.52500m00FA0|=0.90933FAmk^

Solve the above equation to get FA.

FA=230.45Nmk^0.90933mk^=253.42N

Substitute 253.42Nj^ for FA, 231.40Ni^275.78Nj^ for F to determine FD.

FD=231.40Ni^275.78Nj^+253.42Nj^=231.40Ni^22.36Nj^

Write the expression for the magnitude of FD.

FD=(231.40N)2+(22.36N)2=232.48N

Write the expression for the angle made by FD with x-axis.

θ=tan1(22.36N231.40N)=5.5193°

Thus the vertical force at A is 253.42Nj^, and at D is 232.48N making an angle of 5.5193° with the x-axis.

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A worker tries to move a rock by applying a 360-N force to a steel bar as shown. (a) Replace that force with an equivalent force-couple system at D. (b) Two workers attempt to move the same rock by applying a vertical force at A and another force at D . Determine these two forces if they are to be equivalent to the single force of part a.
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Chapter 3 Solutions

Vector Mechanics for Engineers: Statics

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