Question

If a force of magnitude 125 N is exerted an angle of 26◦horizontally as shown, find the force exerted by the hammer claws on the nail. (Assume that the force the hammer exerts on the nail is parallel to the nail).Answer in units of N.

Answer 1

Given,

The horizontal force applied, F=125 N

The angle made by the nail, θ=26°

Distance from the pivot point to point at which the force F is applied, D=28 cm= 28×10⁻² m

The distance from the pivot point to the nail, d=5.86 cm=5.86×10⁻² m

The horizontal force, F, applied to the hammer will give a certain torque to the hammer. Which will transfer the force to the nail.

Thus the torque due to the applied force F will be equal to the torque applied by the hammer to the nail.

Therefore,

`FD=fd\sin \theta`

Where f is the force that the hammer exerts on the nail.

On substituting the known values,

`\begin{gathered} 125*28*10^(-2)=f*5.86*10^(-2)*\sin 26\degree \\ f=(125*28*10^(-2))/(5.86*10^(-2)*\sin 26\degree) \\ =1362.47\text{ N} \end{gathered}`

Thus the force exerted by the hammer on the nail is 1362.47 N