Two blocks with masses m1 and m2 such that m1

Question

Two blocks with masses m1 and m2 such that m1<<m2 are connected by a massless inextensible string and a massless pulley as shown in the figure below. The pulley is rigidly connected to the top of a wedge with angle θ . The coefficient of friction between the blocks is μ . The surface between the lower block and the wedge is frictionless. The goal of this problem is to find the magnitude of the acceleration of each block.
What are the magnitudes of the acceleration of the two blocks? Express your answer in terms of g, μ, m1, m2, and θ

anonymous · Answer

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anonymous · Answer

reply kaisa chahiye?

anonymous · Answer

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anonymous · Answer

@oksuz_ how did u cancel m1a by m1gsin(theta)

anonymous · Answer

because m2>>m1  we can ignore m1 near-by m2.

anonymous · Answer

Okay completely missed that out that's y I couldn't go any further

anonymous · Answer

@oksuz_ but the grader says its wrong

loser66 · Answer

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I just give out my opinion, we can do together. I am a student , too

loser66 · Answer

we have frictionless from the block m2 with the plane
so, $$m_2g + N + T + m_1g = F_2 = m_2 a_2 $$ where g = -9.8 , and N = mgcos $\theta$

anonymous · Answer

yes i can say that tension as they acclerate on both blocks is negative

loser66 · Answer

oh, $m_1gcos \theta$ not just $m_1g$

anonymous · Answer

Tension must be same both side but opposite direction. So we can make equal them each other. However we do not need to use tension, if we calculate forces both side at the same time. We can apply Newton's second law on the whole system one time.

loser66 · Answer

@oksuz_ How can we know it is a system? since $m_1<>a_2$???

anonymous · Answer

well lets try equating sin component of tensional forces and frictional force

anonymous · Answer

if the string is inextensible, both acceleration must be same. Because, they are connected each other.

loser66 · Answer

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anonymous · Answer

This is my solution. But it do not satisfy me any way.

rajat97 · Answer

$$(2mum1gcos \theta+m2gsin \theta-m1gsin \theta)/m1$$ this may be the acceleration of block with mass m1 and $$(mum1gcos \theta+m2gsin \theta)/m2$$ may be the acceleration of block with mass m2

anonymous · Answer

It was one toughie i guess, any way i figured out the right ans

loser66 · Answer

to me, you forgot the normal force from m2 to m1

loser66 · Answer

and the normal force from the plane to m2
My prof taught me that, cannot ignore any force

anonymous · Answer

Normal force is used for finding friction force.

anonymous · Answer

no i didnt i in fact used that to find the frictional force i have clarified in the post of cambridge

loser66 · Answer

ok, may be my knowledge is not enough to discuss further. I am sorry,

anonymous · Answer

no do not underestimate urself it is very easy to understand. do u still have any problem?
@Loser66

loser66 · Answer

my problem is I don't have time. hehehe... a bunch of thing to do for tests next week

anonymous · Answer

g*( (m_2 - m_1)*sin(theta)- 2*mu*m_1*cos(theta) )/(m_2+m_1)