Part one
the figure shows a claw hammer as it pulls a
nail out of a horizontal board. if a force of magnitude 62.5 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 two
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.

Part one : The force exerted  by the hammer claws on the nail is of 460.5 N.

Part two: The force exerted by the surface on the  point of contact with the hammer head is of 431.7 N.

Given data:

The magnitude of horizontal force is, F = 62.5 N.

As per the given diagram, there are four forces on the hammer:

Part One

To find F, sum the moments about the point of contact:

∑τ = Iα

(62.5 N) (30 cm) − (F cos 31°) (4.75 cm) = 0

F = 460.5 N

Thus, we can conclude that the force exerted  by the hammer claws on the nail is of 460.5 N.

Part Two

To find Rₓ and Rᵧ, sum the forces in the x and y directions.

∑Fₓ = ma

62.5 N − F sin 31° + Rₓ = 0

Rₓ = 174.7 N

∑Fᵧ = ma

-F cos 31° + Rᵧ = 0

Rᵧ = 394.7 N

The force exerted by the surface on the  point of contact with the hammer head is due to the net reaction force at point of contact.

Then, the net reaction force at the point of contact is:

Thus, we can conclude that the force exerted by the surface on the point of contact with the hammer head is of 431.7 N.

Learn more about the reaction force here:

link

1) 460.5 N

2) 431.7 N

Explanation:

Draw a free body diagram.  There are four forces on the hammer:

Applied force 62.5 N in the +x direction, 30 cm from the ground

Reaction force Rᵧ in the +y direction, at the point of contact

Reaction force Rₓ in the +x direction, at the point of contact

Reaction force F at 31° from the vertical, 4.75 cm to the left of the point of contact

Part One

To find F, sum the moments about the point of contact:

∑τ = Iα

(62.5 N) (30 cm) − (F cos 31°) (4.75 cm) = 0

F = 460.5 N

Part Two

To find Rₓ and Rᵧ, sum the forces in the x and y directions.

∑Fₓ = ma

62.5 N − F sin 31° + Rₓ = 0

Rₓ = 174.7 N

∑Fᵧ = ma

-F cos 31° + Rᵧ = 0

Rᵧ = 394.7 N

The net reaction force at the point of contact is:

R = √(Rₓ² + Rᵧ²)

R = 431.7 N

wind moving at higher speeds will transfer more energy to the water, resulting in stronger waves.