EBK PHYSICS FOR SCIENTISTS AND ENGINEER
EBK PHYSICS FOR SCIENTISTS AND ENGINEER
6th Edition
ISBN: 9781319321710
Author: Mosca
Publisher: VST
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Chapter 12, Problem 20P

(a)

To determine

The magnitude of the force F applied on the board.

(a)

Expert Solution
Check Mark

Answer to Problem 20P

  181N

Explanation of Solution

Given:

Length of the board, L=3.0 m

Mass of the board, m=5 kg

The angle between the board and the horizontal surface, θ=300

Mass of the block rests on the board, M=60 kg

Distance between the block and the hinge = 80 cm = 0.8 m

Formula used:

Applying rotational equilibrium condition to the board, that is Στ=0 about an axis through the hinge,

  F[Lcosθ]mg[L2cosθ]Mg[L4cosθ]=0

  F=m[L2cosθ]+M[L4cosθ][Lcosθ]g

  F=m(L2)+M(L4)(L)g  (1)

Calculation:

In figure 1,

  mg is the weight of the board

  Fhinge is the force exerted by the hinge

  Mg is the weight of the block

  F is the force acting vertically on the right end of the board

  EBK PHYSICS FOR SCIENTISTS AND ENGINEER, Chapter 12, Problem 20P , additional homework tip  1

FIGURE: 1

Substituting the numerical values in equation (1) ,

  F=5.0 kg × 3m2 + 60 kg × 3 m43 m(9.8 m/s2)

  =5.0 kg ×1.5 m +60 kg × 0.8 m3 m(9.8 m/s2)

  =181N

Conclusion:

The magnitude of the force F applied on the board is 181N .

(b)

To determine

The force exerted by the hinge.

(b)

Expert Solution
Check Mark

Answer to Problem 20P

  456N

Explanation of Solution

Given:

Length of the board, L=3.0 m

Mass of the board, m=5 kg

The angle between the board and the horizontal surface, θ=300

Mass of the block rests on the board, M=60 kg

Distance between the block and the hinge = 80 cm = 0.8 m

Formula used:

Applying ΣFy=0 to the board,

  FhingeMgmg+F=0

  Fhinge=Mg+mgF=(M+m)gF  (2)

Calculation:

Substituting the numerical values in equation (2) ,

  Fhinge=(60kg+5kg)(9.8m/s2)-181NFhinge= 456N

Conclusion:

The force exerted by the hinge is 456N .

(c)

To determine

The magnitude of the force F as well as the force exerted by the hinge, if F is exerted at right angles to the board.

(c)

Expert Solution
Check Mark

Answer to Problem 20P

The magnitude of the force F is 0.16kN

The magnitude of the force exerted by the hinge Fhinge is 0.51kN .

Explanation of Solution

Given:

Length of the board, L=3.0 m

Mass of the board, m=5 kg

The angle between the board and the horizontal surface, θ=300

Mass of the block rests on the board, M=60 kg

Distance between the block and the hinge = 80 cm = 0.8 m

Formula used:

Applying rotational equilibrium condition to the board, that is Στ=0 about the hinge,

  F[L]mg[L2cosθ]Mg[L4cosθ]=0

  F[L]=mg[L2cosθ]+Mg[L4cosθ]

  F=mg[L2cosθ]+Mg[L4cosθ]L=mL2+ML4Lgcosθ  (3)

Applying ΣFy=0 to the board,

  FhingesinθMgmg+Fcos300=0

  Fhingesinθ=(M+m)gFcos300  (4)

Applying ΣFx=0 to the board,

  FhingecosθFsin300=0

  Fhingecosθ=Fsin300  (5)

  Fhinge=Fsin300cosθ  (6)

Dividing equation (4) by equation (5) ,

  FhingesinθFhingecosθ=(M+m)gFcos300Fsin300

  tanθ=(M+m)g+Fcos300Fsin300

  θ=tan1[(M+m)g+Fcos300Fsin300]  (7)

Calculation:

  EBK PHYSICS FOR SCIENTISTS AND ENGINEER, Chapter 12, Problem 20P , additional homework tip  2

FIGURE: 2

Substituting the numerical values in equation (3) ,

  F=5 kg×32+60 kg ×343×9.8 m/s2 cos300F=0.16kN

Substituting numerical values in equation (7) ,

  θ=tan1[(60+5)kg×9.8 m/s2-157 N×cos 300157N×sin 300] = 81.10

Now substituting the numerical values in equation (6) ,

  Fhinge=157N×sin300cos81.10=507.39N= 0.51kN

Conclusion:

The magnitude of the force F is 0.16kN .

The magnitude of the force exerted by the hinge Fhinge is 0.51kN .

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