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When a Force of 6⋅0 N is Exerted at 30° to a Wrench at a Distance of 8 Cm from the Nut, It is Just Able to Loosen the Nut. - Physics

Sum

When a force of 6⋅0 N is exerted at 30° to a wrench at a distance of 8 cm from the nut it is just able to loosen the nut. What force F would be sufficient to loosen it if it acts perpendicularly to the wrench at 16 cm from the nut?

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Solution

In the first case,

\[\tau_1  = 6\sin30^\circ \times \left( \frac{8}{100} \right)\]

In first case,

\[\tau_2  = F \times \left( \frac{16}{100} \right)\]

To loosen the nut, torque in both the cases should be the same.

Thus, we have

\[\tau_1  =  \tau_2\]

\[\Rightarrow F \times \frac{16}{100} = 6\sin30^\circ \times \frac{8}{100}\]

\[\Rightarrow F = \frac{\left( 8 \times 3 \right)}{16} = 1 . 5  N\]

  Is there an error in this question or solution?
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APPEARS IN

HC Verma Class 11, 12 Concepts of Physics 1
Chapter 10 Rotational Mechanics
Q 19 | Page 196
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