Question

Masses, $M_1$ , $M_2$ and $M_3$ are connected by strings of negligible mass which pass over massless and frictionless pulleys $P_1$ and $P_2$ as shown in the figure. The masses move such that the portion of the string between $P_1$ and $P_2$ is parallel to the inclined plane and the portion of the string between $P_2$ and $M_3$ is horizontal. The masses $M_2$ and $M_3$ are $4.0kg$ each and the coefficient of kinetic friction between the masses and the surfaces is $0.25$ . The inclined plane makes an angle of $37^{\circ }$ with the horizontal and the mass $M_1$ moves downwards with a uniform velocity. Find the tension in the horizontal portion of the string.

(Take $g=9.8m{s}^{-2}$ , $sin37^{\circ }$ $\cong$ $3/5$ )

Answer 1

Answer: 9.8

Constant velocity means the net acceleration of the system is zero. Or the net pulling force on the system is zero. While calculating the pulling force, tension forces are not taken into consideration. Therefore,

$M_{1}g=M_{2}gsin37^{\circ }+\mu M_{2}gcos37^{\circ }+\mu M_{3}g$
or $M_1=M_{2}sin37^{\circ }+\mu M_{2}cos37^{\circ }+\mu M_3$
Substituting the values
$M_1=(4)(\frac{3}{5})+(0.25)(4)(\frac{4}{5})+(0.25)(4)=4.2kg$
Since, $M_3$ is moving with uniform velocity
$T=\mu M_{3}g=(0.25)(4)(9.8)=9.8N$