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

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Homework Help Overview

The problem involves calculating the forces exerted by a claw hammer while pulling a nail from a horizontal board, with specific angles and forces provided. The subject area includes mechanics and torque analysis.

Discussion Character

  • Exploratory, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the moment created by the force applied to the hammer and how to calculate the resulting forces. There are attempts to find distances relevant to torque calculations and the effects of angles on the forces involved.

Discussion Status

Some participants have made progress in calculating the first part of the problem, while others are seeking clarification on the second part. There is a mix of approaches being explored, with some guidance provided through hints and equations.

Contextual Notes

Participants are working under the constraints of the problem statement, including specific angles and forces, and are navigating through the calculations without complete consensus on the methods used.

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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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