Part 1) If a force of magnitude 125 N is exerted 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. Part 2) Find the force exerted by the surface on the point of contact with the hammer head. Assume that the force the hammer exerts on the nail is parallel to the nail. Answer in units of N.

Recall, for a body to remain in rotational and translational equilibrium, the net external torque and force acting on a system must be zero. The formula for calculating torque is

Torque = Force x moment arm

Let P be the force exerted by the hammer on the nail. For rotational equilibrium, the torque about the point of contact would be equal. Thus,

For the nail,

distance = 5.86 cm = 5.86/100 = 0.0586 m

moment arm = 0.0586mSin26 = 0.0257

Torque = P x 0.0257 = 0.0257P

For the handle,

Force = 125

perpendicular distance = 28 cm = 28/100 = 0.28

Torque = 125 x 0.28 = 35

By equation both torques,

0.0257P = 35

P = 35/0.0257

P = 1361.87 N

the force exerted by the hammer claws on the nail is 1361.87 N

Part 2)

The force exerted by the surface on the point of contact is the normal force, N

It would be equal to the total downward force. Since the claw is inclined,

Normal force = Downward force x Sinθ

Downward force = P = 1361.87

θ = 26

N = 1361.87Sin26

N = 597 N

the force exerted by the surface on the point of contact with the hammer head is 597 N