Asked by anonymous
A block of mass m rests on a smooth plane of a wedge which is inclined at angle of θ to the horizontal. The wedge is accelerating to the left such that the block does not slide down. Derive an expression for the normal reaction force on the block and the acceleration of the wedge.
Ans given: normal reaction force = mg / cos θ and acceleration = g tan θ.
Ans given: normal reaction force = mg / cos θ and acceleration = g tan θ.
Answers
Answered by
bobpursley
As I read your question, the wedge is moving such that it cancels the downward graviith force on the block.
Downward force on block due to gravity=mg*sinTheta
upward force along the plane due to wedge moving=ma*cosTheta
set them equal
ma*cosTheta=mg*sinTheta
a=g*tanTheta
Downward force: I don't know which force you want here.
Downward force on block due to gravity=mg*sinTheta
upward force along the plane due to wedge moving=ma*cosTheta
set them equal
ma*cosTheta=mg*sinTheta
a=g*tanTheta
Downward force: I don't know which force you want here.
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