Attachment 25586help people please

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- Nov 8th 2012, 07:13 AMabdulrehmanshahforces acting on a slope
Attachment 25586help people please

- Nov 8th 2012, 09:19 AMtopsquarkRe: forces acting on a slope
The begging is not needed and offensive.

You have had some input on question 16 in another thread, so I'll give you an idea of where to start and you can let us know how you did with it.

As in Q16 set up a Free-Body-Diagram (or a force diagram, whatever your instructor may call it.) What you are looking for is a tension in the wire such that m2 has no acceleration on it. That means that m1 has no acceleration either. So set a = 0 in your equations and see what comes up.

-Dan - Nov 8th 2012, 09:33 AMabdulrehmanshahRe: Mechanics
can u please help

- Nov 8th 2012, 10:03 AMebainesRe: Mechanics
I would suggest that you use energy principles: $\displaystyle \Delta KE + \Delta PE = 0$. The kinetic energy of the system is $\displaystyle \frac 1 2 (m_1+m_2)v^2$. The change in potential energy for each mass is equal to mg times the height it rises (note that for m2 since it is moving downhill it's change in PE is negative). So all you need do is come up with an expression for the height rise of each mass, set up the energy equation, and solve for v. Show us your attempt and we'll give more advice if you need it.

- Nov 8th 2012, 10:19 AMabdulrehmanshahRe: Mechanics
0.5(m1+m2)v^2=0.5m1gd-(m2gcos60)s can u please tell me how can there be d inplace of s

- Nov 8th 2012, 10:22 AMabdulrehmanshahRe: Mechanics
i mean what am i doing wrong in order to get the answer required there should be d in place of s and the m2gscos60 is greater than 0.5m1gd

- Nov 9th 2012, 04:22 AMebainesRe: Mechanics
Since s>d, and the two masses are tied together, the system moves distrance 'd' and then m1 hits the pulley, at which point both masses stop. So both masses move distance 'd'. Replace the 's' in your formula by 'd' and solve for v.