Please can anyone describe the motion of a block of mass $m$ sliding on smooth inclined plane of mass $M$ which can slide on the smooth ground. Here, $g$ is the acceleration due to gravity.

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Nsin[theta]=MA
mgsin[theta]=MA
A=mgsin[theta]/M
THEN ACCORDING TO NON INERTIAL FRAME NET ACCELERATION OF m BLOCK IS A+a where a=gsin[theta]
a[net]=mgsin[theta]/M+gsin[theta]
a[net]=gsin[theta][m+M/M]
i think this would be the correct ans for the question

Well, I wish I could help, and also couldn't give you a diagram. First, we must make an agreement to the symbol and the sign, the positive sign and negative sign. Let the block mass m, an inclined plane M, an acceleration for block a, and an acceleration for M is A. We could look the entire objects or it's called a system. The equation will be Sigma Fx=ma, since it's m and a are system we could write as sigma Fx=(m+M)(a cos theta-A). why it should a cos theta? look at the picture, there's an arrow when the small mass are sliding, you could divide it into a cos theta and a sin theta (into a resultant) then you choose the angle is which have a relation to the plane. look again at the equations, it's said Fx, so we should look the acceleration which related to x-axis. then continue to the force, the only force which works on the plane is w sin theta, because there's no other force like friction (because the plane is smooth. so, comes the equation, mg sin theta=(m+M)(a cos theta-A). because you ask the motion of the block, so, a=[mg sin theta/m+M +A]1/cos theta.

If we take wedge frame of reference, then there is a pseudo acceleration on the block of mass m to the right. Then use Normal and 'mg' downwards to find the acceleration.

I cannot paste a diagram here but i hope you understand. Take a normal force $N$ perpendicular to $m$. I will be solving the sum in the frame of reference of ground, so no pseudo force is required.

Let accelerations of $m$ be $a( wrt. M)$ and that of $M$ be $A$.

Note that the acceleration of $m$ wrt. ground in horizontal direction is $(a cosθ - A)$. In vertical direction, it is
$a sinθ.$

i dnt thnk so the horizontal acceleration is acos[tita]-A instead it is acos[tita]+A .as its a non inertial frame so pseudo force acts in opposite direction to the motion of the bigger block

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## Comments

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TopNewestIts depand upon inclination angle

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Transitional

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the best method to solve or to understand this type of motion is psuedo accerlation

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second equation will be : mg = ma sin (theta) + N cos (theta)

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it undergoes a translational and linear motion

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Displacement of centre of mass will be vertically downward.

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Nsin[theta]=MA mgsin[theta]=MA A=mgsin[theta]/M THEN ACCORDING TO NON INERTIAL FRAME NET ACCELERATION OF m BLOCK IS A+a where a=gsin[theta] a[net]=mgsin[theta]/M+gsin[theta] a[net]=gsin[theta][m+M/M] i think this would be the correct ans for the question

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Well, I wish I could help, and also couldn't give you a diagram. First, we must make an agreement to the symbol and the sign, the positive sign and negative sign. Let the block mass m, an inclined plane M, an acceleration for block a, and an acceleration for M is A. We could look the entire objects or it's called a system. The equation will be Sigma Fx=ma, since it's m and a are system we could write as sigma Fx=(m+M)(a cos theta-A). why it should a cos theta? look at the picture, there's an arrow when the small mass are sliding, you could divide it into a cos theta and a sin theta (into a resultant) then you choose the angle is which have a relation to the plane. look again at the equations, it's said Fx, so we should look the acceleration which related to x-axis. then continue to the force, the only force which works on the plane is w sin theta, because there's no other force like friction (because the plane is smooth. so, comes the equation, mg sin theta=(m+M)(a cos theta-A). because you ask the motion of the block, so, a=[mg sin theta/m+M +A]1/cos theta.

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If we take wedge frame of reference, then there is a pseudo acceleration on the block of mass m to the right. Then use Normal and 'mg' downwards to find the acceleration.

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equations as Aveneil mentioned

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gsino. Other wise free fall

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I guess translatory motion

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dude if plane is friction less then its acceleration will a=gsin(thita) so simple

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f=ma+mg

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because of the gravity why its sliding and the shape of the triangle to

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Khan Academy can very well explain you this..

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I cannot paste a diagram here but i hope you understand. Take a normal force $N$ perpendicular to $m$. I will be solving the sum in the frame of reference of ground, so no pseudo force is required.

Let accelerations of $m$ be $a( wrt. M)$ and that of $M$ be $A$.

Note that the acceleration of $m$ wrt. ground in horizontal direction is $(a cosθ - A)$. In vertical direction, it is $a sinθ.$

Writing equations,

$Nsinθ = MA$

$Mg - N cos θ = ma~sinθ$

$Nsinθ = m ( acosθ - A)$

Solving,

$a = \dfrac{(M+m)gsinθ}{M + msin^{2}θ}$

$A = \dfrac{mgsinθcosθ}{M + msin^{2}θ}$

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i dnt thnk so the horizontal acceleration is acos[tita]-A instead it is acos[tita]+A .as its a non inertial frame so pseudo force acts in opposite direction to the motion of the bigger block

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Thank you very much

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if we consider this system wrt wedge then we should take the pseudo force as it is in non inertial frame but you didnt considered it.....why????

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See my solution again, I have mentioned that i am solving it wrt ground frame!

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please can any body help.......

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