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ICSE Class 10 Physics 2014 Paper

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Question : 31 of 66
Marks: +1, -0
Two forces each of 5N5 \text{N} act vertically upwards and downwards respectively on the two ends of a uniform metre rule which is placed at its mid-point as shown in the diagram. Determine the magnitude of the resultant moment of these forces about the midpoint.
Solution:  
Given : F1=F2=5N,d1=d2=0.5mF_1 = F_2 = 5 \text{N}, d_1 = d_2 = 0.5 \text{m}
Torque τ=F×d\tau = F \times d
τ1=F1×d1=5N×0.5m\tau_1 = F_1 \times d_1 = 5 \text{N} \times 0.5 \text{m}
=2.5Nm (Anti-clockwise)= 2.5 \text{Nm} \text{ (Anti-clockwise)}
τ2=F2×d2=5N×0.5m\tau_2 = F_2 \times d_2 = 5 \text{N} \times 0.5 \text{m}
=2.5Nm (Anti-clockwise)= 2.5 \text{Nm} \text{ (Anti-clockwise)}
The resultant moment of force
=Ï„1+Ï„2=2.5+2.5= \tau_1 + \tau_2 = 2.5 + 2.5
=5Nm in Anti clockwise direction.= 5 \text{Nm} \text{ in Anti clockwise direction.}
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