wiz-icon
MyQuestionIcon
MyQuestionIcon
1
You visited us 1 times! Enjoying our articles? Unlock Full Access!
Question

Find the acceleration of the block of mass M in the situation shown in figure (5−E15). All the surfaces are frictionless and the pulleys and the string are light.

Figure

Open in App
Solution


The free-body diagram of the system is shown below:



Let acceleration of the block of mass 2M be a.
So, acceleration of the block of mass M will be 2a.

M(2a) + Mgsinθ − T = 0
⇒ T = 2Ma + Mgsinθ ...(i)
2T + 2Ma − 2Mg = 0
From equation (i),
2(2Ma + Mgsinθ) + 2Ma − 2Mg = 0
4Ma + 2Mgsinθ + 2Ma − Mg = 0
6Ma + 2Mgsin30° + 2Mg = 0
6Ma = Mg
a=g6
Hence, the acceleration of mass M=2a=2×g6=g3 up the plane.

flag
Suggest Corrections
thumbs-up
0
Join BYJU'S Learning Program
similar_icon
Related Videos
thumbnail
lock
Second Law of Motion
PHYSICS
Watch in App
Join BYJU'S Learning Program
CrossIcon