Need Help?

Subscribe to Statics

Subscribe
  • Notes
  • Comments & Questions

Determine the moment of the force about point \(O.\)

\( \sum M_{0}=100\left(\frac{4}{5}\right)(2) \)

\( +100 \left(\frac{3}{5}\right)(5) \)

\( =460 \mathrm{N.m} \)

If the man at  \(B\) exerts a force of  \(P=150 \mathrm{N}\) on his , determine the magnitude of the force  \(\mathbf{F}\)  the man at  \(C\)  must exert to prevent the pole from rotating, i.e., so the  resultant moment about  \(A\) of both forces is zero.

\( \sum M_{A}=0 \)

\( -F\left(\frac{4}{5}\right)(3 \cdot 6)+P \cos (45)(1.8+3.6)=0 \)

\( ∴ F=198 \cdot 8 N \)

In order to pull out the nail at  \(B,\)  the force  \(\mathbf{F}\)  exerted  on the handle of the hammer must produce a clockwise  moment of  \(60 \mathrm{N} \cdot \mathrm{m}\) about point  \(A\) .  Determine the required  magnitude of force  \(\mathbf{F}\) .

\( \sum M_{A}=F \cos 30(0.45) \)

\( +F \sin 30(0.125)=60 \)

Solver\(\longrightarrow\) \( F \cos 30(0.45)+F \sin 30(0.125)=60 \)

\(∴ F=132.7 N \)

No comments yet

Join the conversation

Join Notatee Today!