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1. The problem statement, all variables and given/known data
Consider a wedge of mass ##m## whose upper surface is a quarter circle and is smooth. The lower part is rough. A small block of same mass is kept at the topmost position and the initial position is shown in figure. The block slides down the wedge.
Find the minimum value of coefficient of friction between floor and lower surface of wedge so that the wedge remains stationery during the block's motion on it.
2. Relevant equations
3. The attempt at a solution
I managed to calculate the normal force ##N## exerted by the wedge on the block. This comes out to be ##3mg\sin\theta##. This is also the force exerted by the block on the wedge. However, when I tried to find the normal reaction of the ground on the wedge, Iran into problems.
According to me, we should include the weight of the wedge (##W_W##)+ the vertical component of ##N## + the weight of the block (##W_B## and equate all of them with ##N'## (the normal force acting on the wedge). In other words, $$N'=W_W +N\sin\theta+W_B$$
However, the solution given equates the normal reaction on the wedge by $$N'=W_W+N\sin\theta$$
Could somebody please explain why we ignore the force exerted on the wedge by the block while calculating the normal reaction ##N'## on the wedge? Is it because the weight of the block is a component of the normal force ##N##? If so, how do we know this?
I would be really, really grateful if somebody could kindly explain this in detail to me. Many thanks in anticipation.
Consider a wedge of mass ##m## whose upper surface is a quarter circle and is smooth. The lower part is rough. A small block of same mass is kept at the topmost position and the initial position is shown in figure. The block slides down the wedge.
Find the minimum value of coefficient of friction between floor and lower surface of wedge so that the wedge remains stationery during the block's motion on it.
2. Relevant equations
3. The attempt at a solution
I managed to calculate the normal force ##N## exerted by the wedge on the block. This comes out to be ##3mg\sin\theta##. This is also the force exerted by the block on the wedge. However, when I tried to find the normal reaction of the ground on the wedge, Iran into problems.
According to me, we should include the weight of the wedge (##W_W##)+ the vertical component of ##N## + the weight of the block (##W_B## and equate all of them with ##N'## (the normal force acting on the wedge). In other words, $$N'=W_W +N\sin\theta+W_B$$
However, the solution given equates the normal reaction on the wedge by $$N'=W_W+N\sin\theta$$
Could somebody please explain why we ignore the force exerted on the wedge by the block while calculating the normal reaction ##N'## on the wedge? Is it because the weight of the block is a component of the normal force ##N##? If so, how do we know this?
I would be really, really grateful if somebody could kindly explain this in detail to me. Many thanks in anticipation.
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