where is motion of connected bodies

Motion of connected bodies is an application of the concepts learnt in the chapter of laws of motion. If you can understand that chapter you will be able to do such problems.

Do not hesitate to ask any doubts if you have any.

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Solved Examples Based On Motion Of Connected Bodies Example:

Determine the constant force P that will give the system of bodies shown in Fig. (a) a velocity of 2m/sec after moving 4m from rest. Coefficient of friction between the blocks and the plane is 0.2. Pulleys are smooth.

Solution:

The system of forces acting on connecting bodies is shown in Figure

N1 = 300 N ∴ F1 = N1 = 0.2 × 300 = 60 N

N2 = 1200 cosq = 1200 × 0.6 = 720 N

F2 = 0.2 N2 = 0.2 × 720 = 144 N

N3 = 600N

∴F3 = 0.2 × 600 = 120 N.

Let the constant force to be found be P. Writing work-energy equations to the motion of the system, we get

(PF1 F21200 sin 6 F3) s = W1 + W2 + W3 / 2g (v2 u2)

i.e., (P 60 144 1200 × 0.8 120)4 = 300 + 1200 + 600 / 2 × 9.81 (22 0)

∴ P = 1391.03 N

[Note: Work done is force x distance moved in the direction of force. Hence, N1, N2, N3,(i.e. 300N, 720N, 600N) forces are not contributing to the work done.]

Example: In what distance body A of Fig (a) attains a velocity of 3m/sec after starting from rest? Take u = 0.25. Pulleys are frictionless.

Solution

Let θ1and θ2be the slopes of inclined planes as shown in the figure

sin θ1= 4/5 = 0.8, cos θ1= 0.6

sin q2 = 3/6 = 0.6, cos q2= 0.8 By observing pulley system, it may be concluded that if 1800N block moves a distance 's', 2400N block moves a distance 0.5 s. Hence, if the velocity of 1800 N block is v, that of 2400 N block is 0.5 v. Assuming 2400 N block moves up the plane and 1800 N block moves down the plane, the forces acting on 1800 N that will do work are [Ref. Fig. 14.14(b)] 1800 sin θ1= 1440 N down the plane.

F1 = m 1800 cos θ1= 0.25 × 1800 × 0.6

= 270 N up the plane.

The forces acting on 2400 N block when it slides up are shown in Fig. (c). The forces that do work are

2400 sin θ2= 2400 × 0.6 = 1440 N down the plane

F2 = 0.2 × 2400 cos θ2= 0.2 × 2400 × 0.8

= 384 N down the plane.

Equating work done by various forces to change in the kinetic energy of the system, we get

(1440 270) s (1440 + 384) 0.5 s

= 1800/2 X 9.81 (v2 - 0) + 2400/2 X 9.81 [(0.5 v2) - 0]

Now, v = 3 m/sec

258s = 1800/2.981 X 32 + 2400/2 X 9.81 X 0.25 X 32

Note:

Since 's' is positive, the assumed direction of motion is correct. If it comes out to be negative, recalculations are to be made, since the frictional force changes the sign.

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