Laws of MotionMedium

Question

A trolley of mass 8 kg is standing on a frictionless surface inside which an object of mass 2 kg is suspended. A constant force F starts acting on the trolley as a result of which the string stood at an angle of 37o from the vertical (bob at rest relative to trolley) Then :

Options

A.acceleration of the trolley is 40/3 m/sec2.
B.force applied is 60 N
C.force applied is 75 N
D.tension in the string is 25 N

Solution


F.B.D. of trolley is w.r.t. ground
F.B.D. of suspended mass is w.r.t. Trolley.
T cos 37o − mg = 0     [Equilibrium of mass in y direction w.r.t. trolley]
⇒    T =         T = 25 N
T sin 37o − ma = 0     [Equilibrium of mass in x direction w.r.t. trolley]
⇒    a =         =
F − T sin37 = 8a         [Newton′s II law for trolley in x direction w.r.t. ground]
⇒    F = 8 × + 25 × F = 75 N

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