In this article, i would like to give the concern about the two objects moving in horizontal and vertical directions,
In the below diagram is called free body diagram for the object in horizontal direction. There are three forces are there on the object in horizontal direction, i.e., N,T and m1g.
As the object has no vertical direction, N-m1g=0; N=m1g; the object is moved by cable with tension T with acceleration 'a' i.e.,
T=m1a------1
Next consider the object in vertical direction of mass m2, as the pulley is massless, inextensible and has a fixed length, the object of mass m2 moves with the accleration , therefore
m2g-T=m2a------2
Adding eqn 1 and eqn2, then we get
m2g=(m1+m2)a
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