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a mechanic opens a nut by applying a force of 180 N by using a wrench of length 25 cm. If the mechanic wants to open it by a force of 50N, what should be the length of wrench |
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Answer» 1st CASE, the force applied, F = 120 N and the MAGNITUDE of the perpendicular distance (called radius vector), R=40cm= 40 100 m Torque (τ)=120N× 40 100 =48Nm. In the 2nd case, applied force F = 40 N and the length of the radius vector is to be found. r=? τ=F×r is the same in both the cases. Hence, 48=40×r ∴r=1.2m. Hence, to apply a force of 40 N for the same torque, the mechanic should use a spanner of length 1.2 m. |
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