Given data:
* The force acting on the hammer in the horizontal direction is F = 125 N.
* The distance of horizontal force from the point of contact is d = 28 cm.
* The angle of the nail with the vertical direction is,
[tex]\theta=26^{^{\circ}}[/tex]
* The distance of the nail from the point of contact is x = 5.86 cm.
Solution:
The moment caused by the horizontal force on the hammer is equal to the moment caused by the horizontal component of force at the nail about the point of contact, thus,
[tex]F\times d=F^{\prime}x\text{ sin(}\theta)[/tex]
Substituting the known values,
[tex]\begin{gathered} 125\times28=F^{\prime}\times5.86\times\sin (26^{\circ}) \\ 3500=F^{\prime}\times2.57 \\ F^{\prime}=\frac{3500}{2.57} \\ F^{\prime}=1361.87\text{ N} \end{gathered}[/tex]
Thus, the force exerted on the nail is 1361.87 N or approximately 1362 N.
