Test Index

UPSEE 2019 Solved Paper

Question : 4 of 150

Marks: +1, -0

m

M

Solution:

A free body diagram of a mass pully system is shown below

Now, tension in the string

T - M g = 0

⇒

T = M g

Force in vertical direction,

F_1 = T + m g

\Rightarrow F_1 = M g + m g = (M+m) g

Similarly, force in horizontal direction

F_2 = T = M g

Now, the resultant force

F_{\text{res}} = \sqrt{F_1^2 + F_2^2 + 2 F_1 F_2 \cos 90^{\circ}}

\Rightarrow F_{\text{res}} = \sqrt{F_1^2 + F_2^2}

= \sqrt{(M+m)^2 g^2 + M^2 g^2}

\Rightarrow F_{\text{res}} = g \sqrt{(M+m)^2 + M^2}

Hence, the resultant force is

g \sqrt{(M+m)^2 + M^2}

Go to Question:

Prev Question

Next Question