Physics, asked by kneerajabtech, 11 months ago

A man of mass m is inside a box of same mass .The whole system is suspended with ideal strings and pulley as shown in above figure.The tension t1 in the string so as to keep the system in equilibrium is​

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Answered by qwchair
5

Answer:The tension t1 in the string so as to keep the system in equilibrium is​ mg

Explanation: As per the figure, the net force on the top most point is 2mg as the weight of both the man and box is same. So to keep entire system in balance the tension on T1 is half of the total force. That is mg.

Answered by madeducators4
13

Given :

Mass of man :

= m

Mass of box :

=m

To Find :

Tension T_{1} in the string so as to keep the system in equilibrium =?

Solution :

Magnitude of weight of man and box acting downward :

= mg

Here since the 2nd pulley is an ideal pulley so the force acting on it will be zero , so that :

2T_{2}= T_{1}

Now for equilibrium upward force should balance the downward , so :

T_{1} +T_{2} =mg+ mg - T_{2}\\T_{1}+2T_{2}= 2mg\\2T_{1}=2mg\\T_{1}=mg

Therefore tensionT_{1} in the string is mg.

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