How do you calculate the forces involved when a claw hammer pulls out a nail?

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To calculate the forces involved when a claw hammer pulls out a nail, the moment at the point of contact is determined using the equation M = Fd, where a horizontal force of 145 N is applied. The first part of the solution yields a force exerted by the hammer claws on the nail of approximately 0.96 kN at an angle of 64.9 degrees above horizontal. For the second part, equilibrium equations ΣFx = 0 and ΣFy = 0 are used to find the reaction force at the contact point, resulting in a value of 908.75 N. The calculations demonstrate the relationship between applied forces, angles, and resulting moments in this scenario. Understanding these forces is crucial for analyzing the mechanics of using a claw hammer effectively.
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Homework Statement


The figure below shows a claw hammer as it is being used to pull a nail out of a horizontal board where θ = 25.1°. A force of 145 N is exerted horizontally as shown. (Assume that the force the hammer exerts on the nail is parallel to the nail.)

(a) Find the force exerted by the hammer claws on the nail.
(b) Find the force exerted by the surface on the point of contact with the hammer head.

Homework Equations


Moment M=Fd

The Attempt at a Solution


The moment at point of contact would be M = 145N * 30cm
I don't know how to proceed from here.. Any help would be appreciated.
 

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Find the distance from the point of contact to where the hammer grasps the head of the nail. The nail acting on that lever exerts a counterclockwise torque.
 
Stephen Tashi said:
Find the distance from the point of contact to where the hammer grasps the head of the nail. The nail acting on that lever exerts a counterclockwise torque.

Ok I have calculated 1st part:
M=145*0.3
Fcos\vartheta=M/0.05
=> F=0.96 kN 64.9 degrees above horizontal.

What about the second part?
 
Last edited:
Its ok, I figured it out.. Thank you for your hint.. :)

\SigmaFx = 0
\SigmaFy = 0
Using these equations I got reaction force as 908.75 N.
 
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