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If I apply a force perpendicular to the side of a 2D block at $\tfrac{H}{2}$ its height with a fixed bottom corner opposite the applied force, and an anchor bolt is placed $\tfrac{W}{2}$ from the pivot. How much force will be applied to the anchor bolt relative to the initial applied force?

  • Block Mass : M
  • Block Width : W
  • Block Height : H
  • Applied Force : F
  • Distance between force and pivot : r

I think the torque relative to the pivot will be: $$ F \cdot r \cdot \sin( ∠A ) $$

I"m unsure how much of that torque is applied to the anchor bolt. Would it be 50% of the torque at the pivot since it is in the bottom middle of the block?

  • $\begingroup$ Is the anchor bolt in the middle ($\tfrac{W}{2}$ distance from pivot?) $\endgroup$ – ja72 Aug 28 '17 at 19:40
  • $\begingroup$ Yes, it's $ \tfrac{W}{2} $ $\endgroup$ – Cggart Aug 28 '17 at 19:41

You need to do a free body diagram ( I included the weight $W=mg$)


Then balance the forces and moments about the pivot

$$\begin{align} -F +A_x & = 0 & \mbox{x-axis} \\ -m g -A_y + B_y & = 0 & \mbox{y-axis} \\ \frac{W}{2} (B_y-m g) + \frac{H}{2} F & = 0 & \mbox{torque} \end{align} $$

to be solved for $A_x$, $A_y$ and bold froce $B_y$.

  • $\begingroup$ Yes. The arbitray convention I used would result in negative bolt force. Any negative result needs the sense of the force flipped from the diagram. $\endgroup$ – ja72 Aug 28 '17 at 20:27
  • $\begingroup$ @sammygerbil - AHA, now I see. I will correct the equations. Thank you for checking my work. The moment of $F$ is positive (CCW from pivot). $\endgroup$ – ja72 Aug 29 '17 at 1:09

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