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Magazine Chapter 9: Q23 PE (page 317)
What force does the nail puller inExercise\({\rm{9}}{\rm{.19}}\)exert on the supporting surface? The nail puller has a mass of\({\rm{2}}{\rm{.10}}\;{\rm{kg}}\).
Short Answer
The force exerted by the puller on the supporting surface is \({\rm{1}}{\rm{.30 \times 1}}{{\rm{0}}^{\rm{3}}}\;{\rm{N}}\).
Step by step solution
Definition of the force
A force is an external factor that can change the rest or motion of a body. It has a size and a general direction.
Free-body diagram of the system
The diagram is shown below:
Here,\({I_i} = 45\;{\rm{cm}}\),\({I_o} = 1.80\;{\rm{cm}}\) and \(F = 50\;{\rm{N}}\) .
Calculation of the force
The torque about the pivot point is zero in equilibrium. So,
\(\begin{align}N{l_o} &= \left( {{l_o} + {l_i}} \right){F_i}\\N &= \frac{{\left( {{l_o} + {l_i}} \right){F_i}}}{{{l_o}}}\\N &= \left( {1 + \frac{{{l_i}}}{{{l_o}}}} \right){F_i}\end{align}\)
For the given values,
\(\begin{align}N &= \left( {1 + \frac{{45}}{{1.80}}} \right) \times 50\;N\\ &= 1.30 \times {10^3}\;N\end{align}\)
Thus, the force is \({\rm{1}}{\rm{.30 \times 1}}{{\rm{0}}^{\rm{3}}}\;{\rm{N}}\).
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