Full Question
The figure shows a claw hammer as it pulls a nail out of a horizontal board.
If a force of magnitude 181 N is exerted horizontally as shown, 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.
See attachment for figures
Answer:
1026.85 Newton
Explanation:
Given
Height (h)= 29 cm,
ℓ= 5.57 cm,
α= 32◦,and
F= 181
Let f = the horizontal force exerted by the surface at the point of contact on the hammer.
Let R = Reaction force of the nail at the point of contact with the hammer
In Newton’s second law,
We have our horizontal sum to be:
Fx=f + F- Rsinα= 0 ----- Make f the subject of formula
f = Rsinα - F
Sum of the torques about the point of contact of the hammer with the table,
τ=R ℓcosθ-F
Where τ= 0
So, 0 = R ℓcosθ-Fh
R = Fh/ ℓcosθ
R = 181 * 29/(5.57 * cos32)
R = 5249/4.723627895591292
R = 1111.22N
So, f = 1111.22sin32 - 181
f = 407.86
Also in Newton's Second Law,
We have our vertical sum to be:
Fy=n-Rcosα= 0----- make n the subject of formula
n = Rcosα
n = 1111.22 * cos32
n = 942.37
Using F² = f² + n²
F = √f² + n²
F = √407.86² + 942.37²
F = 1026.85N
