0
$\begingroup$

enter image description here

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?

$\endgroup$
  • $\begingroup$ 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
1
$\begingroup$

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

fbd

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 $B_y$.

$\endgroup$
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.