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all the surfaces are frictionless...... now as it is clear that as the bigger block move to right with an acceleration the smaller block will move down and towards right with the same accleration....now if we draw the FBD of the smaller block we will see that there is no force directed towards the right to it...so how does the smaller block move towards right...also the string remains vertical all the time soo there is no component of force making it to move rightward......
Comments (5)
after small time t, the cos component of tention will make the smaller back move towards right.
For the given figure, we are required to talk about the dynamics part only, the acceleration of small and big block.
i am not really convinced with the answers...... if the answer u told is right then how come we solve all other questions of mechanics having similar setup without taking the component of tension... if your approach is right then can u please find out the acceleration of smaller block.....
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http://www.goiit.com/posts/list/mechanics-intresting-question-98803.htm
The answer of the question came out to be zero. The contact will be lost and the right block will move towards the right. But initially there is no hor. force on the left block. As the right block moves right, the pulley also moves right and the string will become angled. the tension in the string will provide the horizontal force necessary for the block to move right. the normal force will be zero as contact is lost.