Where did I go wrong in finding the length x using the grabber-statics example?

mrshlnnxfa · Feb 10, 2017

Hello,

We should get the length x , when the grabber can just hold the box.

Diagram of a gripper with two crossed arms pivoting at a central joint, each arm carrying force G/2 upward, gripping a green box of weight G with friction coefficient mu0; lengths x, l1, l2, angle alpha and width b labelled.

I made two moment equations:

$e[czech]{babel}%20M_{1}=\frac{G}{2\mu%20}*\sqrt{(l_{1}+l_{2})^{2}+(\frac{b}{2})^{2}}\end{align*}.gif$

$yle\begin{align*}\usepackage[czech]{babel}%20M_{2}=\frac{G}{2}*\sin%20\alpha%20*%20x\end{align*}.gif$

Then I put them together into one equation:

$Equation M1 = M2 written as plain mathematics, showing two moment terms set equal to each other.$

And from that equation I found out x:

$Formula: x = csc α * sqrt(b^2 + 4(l1 + l2)^2) / (2μ)$

Unfortunately the result seems to be wrong. Can you please help me? Where I made the mistake?

akshay86 · Feb 11, 2017

i think the frictional force balance the wight of the box.we have to give the normal force on the box that give rise to the friction by giving a torque to the grabber

mrshlnnxfa · Feb 11, 2017

I finally got it. Here is the solution.

Two typeset equations: (x sin α)(½G) + ½G(½b) = G/(2μ0)(l1 + l2), and below it x = (2(l1 + l2) − μ0 b) / (2μ0 sin α).

akshay86 · Feb 12, 2017

Yes,that is the correct answer

Where did I go wrong in finding the length x using the grabber-statics example?

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