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Question: In the system shown, pulley and string are ideal. The vertical upward pull F is being increased grad...

In the system shown, pulley and string are ideal. The vertical upward pull F is being increased gradually. Find the magnitude of F and acceleration of the 5  kg5\;kg block at the moment the 10  kg10\;kg leaves the floor.

Explanation

Solution

We can assume the pulley-mass to be a system. There are two forces which act on the system. One is the tension on the string, and the other is the force due to gravitation. Then the resultant vector is the force acting on the whole system.

Formula used: T=mgT=mg

Complete step by step answer:
We know that a pulley is a simple device that can lift heavy objects with minimum energy. Here a block of some mass is attached to the pulley. A pulley consists of a tensile string or rope, which is used to tie and lift the heavy body and a shaft above which the string is kept. The shaft is circular generally and allows the easy movement of the string.
The force needed to lift the massive body is greater than the force needed to pull it using a pulley. Hence pulleys are used to lift heavy objects
Consider the pulley-mass system as shown in the figure. Where M=10kgM=10kg and m=5kgm=5kg is the mass of the blocks. Let TT be the tension on the string and FF be the force due to gravitation.

Clearly, we can say that F=2TF=2T
    T=F2\implies T=\dfrac{F}{2}
We also know that T=mgT=mg
Just as the 10  kg10\;kg block leaves the floor, the tension experienced by the string is given as T=10×10=100NT=10\times 10=100N
Clearly, F=2×100=200NF=2\times 100=200N
At the same instant, let us assume that the block of mass m=5kgm=5kg is accelerated in the upward direction, say aa.
Clearly, the total force on the block is given as Tmg=FT-mg=F
And also , F=maF=ma
Then, we can say that Tmg=maT-mg=ma
Substituting the values we have, 10050=5×a100-50=5\times a
    a=10m/s2\implies a=10m/s^{2}

Thus the net force FF acting in the upward direction is 200  N200\;N and the acceleration of the body m=5kgm=5kg is 10m/s210m/s^{2}

Note: Since tension and force are both vectors, the direction of the resultant force will be the addition of the two vectors using the parallelogram law of addition. Here only the force on the pulley is asked, hence the magnitude is sufficient.