Question

A claw hammer is used to pull a nail out of a board. The nail is at an angle of 60^o to the board, and a force ${\overrightarrow{\mathbf{F}}}_{\mathbf{1}}$of magnitude 400 N applied to the nail is required to pull it from the board. The hammerhead contacts the board at a point A, which is 0.800 m from where the nail enters the board. A horizontal force ${\overrightarrow{\mathbf{F}}}_{\mathbf{2}}$is applied to the hammer handle at a distance of 0.300 m above the board. What magnitude of force ${\overrightarrow{\mathbf{F}}}_{\mathbf{2}}$is required to apply the required force (${\overrightarrow{\mathbf{F}}}_{\mathbf{1}}$) to the nail? (Ignore the weight of the hammer.)

Short answer

The magnitude of F₂ is 92.4 N.

Step-by-step solution

The given data

Given that a claw hammer is used to pull a nail out of a board. The nail is at an angle of 60^o to the board, and a force${\overrightarrow{F}}_1$ of magnitude 400 N applied to the nail is required to pull it from the board. The hammer head contacts the board at a point A, which is 0.800 m from where the nail enters the board. A horizontal force${\overrightarrow{F}}_2$ is applied to the hammer handle at a distance of 0.300 m above the board.

Force $F_1=400$N

Formula used

Torque$\mathit{\tau }\mathbf{=}\mathit{F}\mathit{l}$

Where Fis force exerted and l is distance.

Draw a free-body diagram of the beam

The moment arm for the force F₁ is $\left(0.08m\right)\sin 60°$.

The moment of arm for F₂ is 0.3 m.

Applying equilibrium condition on torque

This gives torque$\tau =0$

Therefore,

$$\begin{gathered} F_2\left(0.3m\right)=F_1\left(0.8m\right)\sin 60° \\ {F}_2=\left(400N\right)\left(0.231\right)=92.4N \end{gathered}$$

So the magnitude of F₂ is 92.4 N