
Concept explainers
The force exerted by the two personson the board.

Answer to Problem 76A
The force exerted by the person, near the box, on the board is 62.8 N and the force exerted by the other person on the board is 37.65 N .
Explanation of Solution
Given:
The mass of the wooden board is m=4.25 kg .
The length of the wooden board is 1.75 m .
The mass of the box is m=6.00 kg .
Formula used:
The torque is the product of the lever arm and the applied force.
The expression for the torque as follows:
τ=Frsinθ
Here, F is the applied force, r is the radius of the wheel, and θ is the angle made by anobject with the vertical.
Calculation:
Consider the force exerted by the person A on the board is FA .
Consider the force exerted by the person B,near the box, on the board is FB .
Consider the force due to the weight of the box is denoted by Fg .
Consider the force due to the centre of gravity is denoted by Fcog .
Consider a crank is connected to a chained wheel.
The forces acting on the board are shown in the Figure below.
For the equilibrium of the forces acting on the board,
FA+FB−Fg−Fcog=0FA+FB=Fg+FcogFA+FB=(mboardg)+(mboxg)FA+FB=(mboard+mbox)gFA+FB=(4.25 kg+6.00 kg)×9.80 m/s2FA+FB=100.45 kg⋅m/s2×(1 N1 kg⋅m/s2)FA+FB=100.45 N
Consider the torque exerted on the board due to the forces FA,FB,Fg,Fcog are denoted by
τA,τB,τg,τcog .
For Equilibrium of the torque acting on the board,
τA+τB=τg+τcog0+τB=τg+τcogτB=τg+τcogFBrB=[Fgrg]+[Fcogrcog]FB(1.25 m+0.50 m)=[41.65 N×1.25 m]+[41.65 N×1.25 m]FB(1.75 m)=[36.44 N⋅m]+[73.5 N⋅m]FB=(109.94 N⋅m1.75 m)FB=62.8 N
The force exerted by the person A on the board is
FA+FB=100.45 NFA+62.8 N=100.45 NFA=37.65 N
Conclusion:
Therefore, the force exerted by the person, near the box, on the board is 62.8 N and the force exerted by the other person on the board is 37.65 N .
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