anonymous
  • anonymous
Two boxes (m1 = 43.0 kg and m2 = 39.5 kg) are connected by a light string that passes over a light, frictionless pulley. One box rests on a frictionless ramp that rises at 30.0^\circ above the horizontal (see the figure below), and the system is released from rest. find the acceleration of each box
Physics
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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anonymous
  • anonymous
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anonymous
  • anonymous
this is what i came up with T-mg =ma t-mgsibthera=ma
anonymous
  • anonymous
Both masses have the same acceleration: \[a=\frac{M-m\sin 30^\circ}{M+m}g\] where m is the mass on the incline.

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anonymous
  • anonymous
If we realize that the tension acts parallel to the surface of the incline, we can sum the forces that act parallel to the surface of the incline when we analyze \(m_1\). Let's sum the forces about \(m_1\) parallel to the incline. \[F = ma \rightarrow m_1 a = T - m_1 g \sin(\theta)\]Now, let's sum the vertical forces about \(m_2\). \[-m_2 a = T - m_2 g\] T and a must be equal for both masses. Since we want to know a, let's solve both for T and set them equal to each other. \[T = m_1 a + m_1 g \sin(\theta)\]\[T = -m_2 a + m_2 g\]\[m_1 a + m_1 g \sin(\theta) = m_2 a + m_2 g\]We can now solve for a\[a (m_1 + m_2) = m_2g - m_1 g \sin(\theta)\]\[a = {m_2 g - m_1 g \sin(\theta) \over m_1 + m_2}\]
anonymous
  • anonymous
let the acceleration of the first block m1 be a hence the acceleration of second block also will be a due to both connected by same string. |dw:1332736369599:dw| hence from the figure:: m1gsin30-T=m1a...(1) T is tension T-m2g=m2a....(ii) from (I) and(ii) (m1/2-m2)g=(m1+m2)a a=(m1/2-m2)g/(m1+m2).. it will give u negative value of a . that means ,which direction we have assumed,opposite to it acceleration direction will be..
anonymous
  • anonymous
eashmore why is m2a negative in the equation -m2a=T-m2g
anonymous
  • anonymous
Such that the acceleration of m1 has the same direction as m2. The sign in wrong in the third to last equation. \[m_1 a + m_1 g \sin(\theta) = -m_2 a + mg\]
anonymous
  • anonymous
srry but i dont see why its negative
anonymous
  • anonymous
Let's just assume that m_2 accelerates downwards, therefore m_1 will accelerate up the incline. Additionally, T will act up the incline on m_1 and upwards on m_2. \[m_1 a = T - m_2 g \sin(\theta)\]\[m_2 a = -T + m_2 g \rightarrow -m_2 a = T - m_2 g\]
anonymous
  • anonymous
so as one mass climbs up the incline the other is being pulled down
anonymous
  • anonymous
That is a necessary condition here. Both blocks must move in the same direction. This is because they are linked by an inextensible wire.
anonymous
  • anonymous
so what i said is correct
anonymous
  • anonymous
Yes. If the block moves up the incline, the other must move downwards. The vise-versa case is also valid.
anonymous
  • anonymous
also under what condition do i set the total force equal to ma
anonymous
  • anonymous
When the objects are in motion. Remember that\[\sum F = ma \]
anonymous
  • anonymous
in this situation we used ma but in the question we r told that the boxes are at rest
anonymous
  • anonymous
The boxes are released from rest.... and we are explicitly asked to find the acceleration of the boxes. Therefore, the boxes must move by the problem statement.
anonymous
  • anonymous
okay and also when do we replace w with mg
anonymous
  • anonymous
Anytime. Weight is almost always equal to mg.

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