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Let " ${T}_{1}$ " and " ${T}_{2}$ " be the tensions in the two strings as shown in the figure.
As the pulleys are “mass – less”, we have following force relations,
$$\Rightarrow {T}_{1}=\frac{600}{2}=300N$$
$$\Rightarrow {T}_{2}=\frac{{T}_{1}}{2}=\frac{300}{2}=150N$$
On the other hand, weights of the blocks are 100 N and 200 N. Since tension in the string is 150 N, only block of 10 kg with weight 100 N will move up. The other block will remain stationary. Force analysis on the mass of 10 kg in shown on the right hand side of the figure shown above. The acceleration of the block of mass 10 kg is :
$$\Rightarrow a=\frac{{T}_{2}-mg}{m}=\frac{150-10X10}{10}=5\phantom{\rule{1em}{0ex}}m/{s}^{2}$$
Problem 3 : Two blocks of mass 1 kg and 2 kg are connected with a string that passes over a “mass – less” pulley. A time dependent force "F = 20 t" acts on the pulley as shown. Find the time when the block of mass 1 kg looses contact from the floor.
Solution : The block of mass 1 kg is about to be lifted when tension in the string is equal to the weight of the block such that normal force between the surfaces becomes zero. This means that :
$$\Rightarrow T=mg=1X10=10N$$
On the other hand, the analysis of force on the pulley, "A", yields,
$$\Rightarrow F=2T$$
$$\Rightarrow T=\frac{F}{2}=\frac{20t}{2}=10t$$
Equating two equations, we have :
$$\Rightarrow 10t=10$$
$$t=1\phantom{\rule{1em}{0ex}}s$$
Problem 4 : In the arrangement shown in the figure, the block “A” moves with a velocity 4 m/s towards right. The string and the pulleys are “mass-less” and friction is absent everywhere. What is relative velocity of block “B” with respect to “A”?
Solution : The blocks are connected by a single string, which is intervened by two pulleys. We can connect velocities of blocks, if we have constraint relation for the length of string. First differentiation of the constraint relation will yield the required relation between velocities.
Since the pulleys are connected to blocks as their part (their relative positions do not change due to motion), we treat block and the attached pulley as a single entity, whose positions are same. Following the procedure for writing constraint relation, we select the fixed end of the string as the reference point.
From the figure,
$${x}_{A}+\left({x}_{A}+{x}_{B}\right)+\left({x}_{A}+{x}_{B}\right)=L$$
$$\Rightarrow 3{x}_{A}+2{x}_{B}=0$$
Differentiating w.r.t time, we have :
$$\Rightarrow 3\frac{\u0111{x}_{A}}{\u0111t}+2\frac{\u0111{x}_{B}}{\u0111t}=0$$
There is an important subtle point here. The positions of blocks are on either side of the reference point (not on the same side as usually is the case). If positive direction of reference x-direction is towards right as shown in the figure, then velocities of two blocks are :
$${v}_{A}=\frac{\u0111{x}_{A}}{\u0111t}$$
$${v}_{B}=-\frac{\u0111{x}_{B}}{\u0111t}$$
Substituting in the equation, we have :
$$\Rightarrow 3{v}_{A}-2{v}_{B}=0$$
$$\Rightarrow {v}_{B}=\frac{3{v}_{A}}{2}$$
The sign of this relation (positive) denotes that velocities of blocks "A" and "B" are both in the reference direction of x-axis.
Relative velocity of “B” with respect to “A” is :
$${v}_{BA}={v}_{B}-{v}_{A}=\frac{3{v}_{A}}{2}-{v}_{A}$$
$$\Rightarrow {v}_{BA}=\frac{{v}_{A}}{2}=\frac{4}{2}=2\phantom{\rule{1em}{0ex}}m/s\phantom{\rule{1em}{0ex}}\text{towards right}$$
Problem 5 : A ball and rod of mass “m” and “nm” respectively are held in positions as shown in the figure. The friction is negligible everywhere and mass of the pulleys and the strings are also negligible. At a certain time, the ball and rod are released simultaneously. If the length of the rod is “L”, what time would the ball take to reach the other end of the rod. Consider "n>1".
Solution : The ball and the moving pulley are connected through the same string that goes over a fixed pulley. As such, their accelerations are same. Now, the rod is hanging from a string that passes over the moving pulley and its other end is attached to a fixed point. From the constrain relation as worked out in the subject module Pulleys , we know that acceleration of the rod is twice that of the moving pulley. Hence, we can conclude that acceleration of the rod is twice as that of the ball. Let us, then, consider that ball and rod move with accelerations “a” and “2a” respectively.
As the rod moves down (n>1), the ball moves up. The motions are in opposite direction. Since two elements are moving in opposite directions, the relative acceleration is arithmetic sum “a + 2a = 3a”. Now, using equation of motion for constant acceleration, we have :
$$L=\frac{1}{2}\left(3a\right){t}^{2}$$
$$\Rightarrow T=\sqrt{\left(\frac{2L}{3a}\right)}$$
In order to evaluate this expression, we need to know acceleration “a”. From the free body diagram of the moving pulley, we note that the tension in the string attached to ball is twice the tension in the string attached to the rod.
The free body diagram of the ball and the rod are shown on the right side of the above figure. The force analysis in the vertical directions yields following relations :
$$nmg-T=nmX2a$$
$$2T-mg=ma$$
Multiplying first equation by “2” and eliminating “T”, we have :
$$\Rightarrow 2nmg-mg=4nma+ma=\left(4n+1\right)ma$$
$$a=\frac{\left(2n-1\right)g}{\left(4n+1\right)}$$
Putting this value in the expression of time, we have :
$$t=\sqrt{\left(\frac{2L\left(4n+1\right)}{3\left(2n-1\right)g}\right)}$$
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