Vector Mechanics for Engineers: Statics, 11th Edition
Vector Mechanics for Engineers: Statics, 11th Edition
11th Edition
ISBN: 9780077687304
Author: Ferdinand P. Beer, E. Russell Johnston Jr., David Mazurek
Publisher: McGraw-Hill Education
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Chapter 3.1, Problem 3.5P

A 300-N force is applied at A as shown. Determine (a) the moment of the 300-N force about D, (b) the magnitude and sense of the horizontal force applied at C that creates the same moment about D, (c) the smallest force applied at C that creates the same moment about D.

(a)

Expert Solution
Check Mark
To determine

Moment of 300N force about D.

Answer to Problem 3.5P

Moment of 300N force about D is 41.7Nm in counter clockwise direction.

Explanation of Solution

Refer Figure 1.

Vector Mechanics for Engineers: Statics, 11th Edition, Chapter 3.1, Problem 3.5P , additional homework tip  1

Write an expression to calculate the force vector acting at point A

F=(Fcosθ)i+(Fsinθ)j (I)

Here, F is the magnitude of force acting at point A, θ is the angle between the force vector and AB.

Write an expression to calculate the position vector along line connecting A and D.

r=(AO)i(AE)j (II)

Here, r is the position vector along line connecting A and D, AO is the horizontal distance between A and D, AE is the vertical distance between A and D.

Write an expression to calculate the moment about point D.

MD=r×F (III)

Here, MD is the moment about point D.

Substitute equation (I) and (II) in equation (III) to rewrite equation (III).

MD=((AO)i(AE)j)×((Fcosθ)i+(Fsinθ)j) (IV)

Conclusion:

Substitute 300N for F , 25° for θ, 100mm for AO and 200mm for AE in equation (IV) to find MB.

MD=((((100mm)(1m1000mm))i((200mm)(1m1000mm))j)×(((300N)cos(25°))i+((300N)sin(25°))j))=((0.001m)i(0.002m)j)×((271.89N)i+(126.785N)j)=(41.7Nm)k

Therefore, the of 300N force about D is 41.7Nm in counter clockwise direction.

(b)

Expert Solution
Check Mark
To determine

The magnitude and sense of horizontal force applies at C that creates the same moment about D.

Answer to Problem 3.5P

The magnitude and sense of horizontal force applies at C that creates the same moment about D is 333.60N.

Explanation of Solution

Refer Figure 2.

Vector Mechanics for Engineers: Statics, 11th Edition, Chapter 3.1, Problem 3.5P , additional homework tip  2

Write an expression to calculate the position vector along line connecting C and D.

rCD=(OB)i(CF)j (V)

Here, rC/D is the position vector D, BO is the horizontal distance between C and D, CF is the vertical distance between C and D.

Write an expression to calculate the moment about point D.

MD=rC/D×C (VI)

Here, C is the force acting at point C.

Write an expression to calculate the force acting at point C.

C=Ci (VII)

Here, C is the magnitude of force acting at point C.

Substitute equation (V) and (VII) in equation (VI) to rewrite equation (VI).

MD=((OB)i(CF)j)×(Ci)=((C)(CF))(k) (VIII)

Write an expression to determine the magnitude of the moment about D.

MD=(C)(CF) (IX)

Here, MD is the magnitude of moment about D.

Rearrange the equation (IX) to determine the magnitude of force acting at point C.

C=MDCF (X)

Conclusion:

Substitute 41.7Nm for MD and 125mm for CF in equation (X) to find C.

C=41.7Nm(125mm)(1m1000mm)=333.60N

Therefore, the magnitude and sense of horizontal force applies at C that creates the same moment about D is 333.60N.

(c)

Expert Solution
Check Mark
To determine

The smallest force applied at C that creates the same moment about D.

Answer to Problem 3.5P

The smallest force applied at C that create the same moment about D is 176.8N at an angle 58° before horizontal.

Explanation of Solution

Refer Figure 3.

Vector Mechanics for Engineers: Statics, 11th Edition, Chapter 3.1, Problem 3.5P , additional homework tip  3

For minimum force, the force vector Cmin   will be perpendicular to the line DC.

Write an expression to calculate the smallest magnitude of force vector Cmin that creates the same moment about D.

Cmin=MDDCsinθ (XI)

Here, Cmin is the smallest magnitude of force vector Cmin that creates the same moment about D, θ is the perpendicular angle between DC and force vector Cmin.

Write an expression to calculate the length of DB.

DC=(OB)2+(CF)2 (XII)

Substitute equation (XII) in (XI) to rewrite the equation (XI).

Q=MD(OB)2+(CF)2sinθ (XIII)

Write an expression to calculate the angel α.

α=tan1(CFOB) (XIV)

Write an expression to calculate the direction of force.

ϕ=θα (XV)

Substitute equation (XIV) in (XV) to rewrite find (XV).

ϕ=θtan1(CFOB)

Conclusion:

Substitute 41.7Nm for MD, 90° for θ, 200mm for OB and 125mm for CF in equation (XIII) to find Cmin.

Cmin=41.7Nm((200mm)(1m1000mm))2+((125mm)(1m1000mm))2sin(90°)=176.8N

Write an expression to calculate the angel α.

α=tan1(CFOB) (XIV)

Write an expression to calculate the direction of force.

ϕ=θα (XV)

Substitute equation (XIV) in (XV) to rewrite find (XV).

ϕ=θtan1(CFOB)

Substitute 90° for θ, 200mm for OB and 125mm for CF in equation (XV) to find ϕ.

ϕ=90°tan1((125mm)(1m1000mm)(200mm)(1m1000mm))=58°

Therefore, the smallest force applied at C that create the same moment about D is 176.8N at an angle 58° before horizontal.

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Chapter 3 Solutions

Vector Mechanics for Engineers: Statics, 11th Edition

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