OpenStudy (anonymous):

Two blocks with masses m1 and m2 such that m1< 4 years ago

OpenStudy (anonymous):

4 years ago
OpenStudy (anonymous):

4 years ago
OpenStudy (anonymous):

4 years ago
OpenStudy (anonymous):

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

4 years ago
OpenStudy (anonymous):

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

4 years ago
OpenStudy (anonymous):

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

4 years ago
OpenStudy (anonymous):

@oksuz_ but the grader says its wrong

4 years ago
OpenStudy (loser66):

|dw:1380457082999:dw| I just give out my opinion, we can do together. I am a student , too

4 years ago
OpenStudy (loser66):

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 $$\theta$$

4 years ago
OpenStudy (anonymous):

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

4 years ago
OpenStudy (loser66):

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

4 years ago
OpenStudy (anonymous):

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.

4 years ago
OpenStudy (loser66):

@oksuz_ How can we know it is a system? since $$m_1<<m_2$$ can we assume that $$a_1>>a_2$$???

4 years ago
OpenStudy (anonymous):

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

4 years ago
OpenStudy (anonymous):

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

4 years ago
OpenStudy (loser66):

4 years ago
OpenStudy (anonymous):

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

4 years ago
OpenStudy (rajat97):

$(2mum1gcos \theta+m2gsin \theta-m1gsin \theta)/m1$ 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

4 years ago
OpenStudy (anonymous):

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

4 years ago
OpenStudy (loser66):

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

4 years ago
OpenStudy (loser66):

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

4 years ago
OpenStudy (anonymous):

Normal force is used for finding friction force.

4 years ago
OpenStudy (anonymous):

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

4 years ago
OpenStudy (loser66):

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

4 years ago
OpenStudy (anonymous):

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

4 years ago
OpenStudy (loser66):

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

4 years ago
OpenStudy (anonymous):

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

4 years ago