Skip to main content

Visit the website manishverma.site for latest posts, courses, admission & more.

Pulley + Block + Rope

A uniform rope of linear mass density $\lambda $ is used to release block m with uniform acceleration a. Find the tension at a point P on the rope at a distance l from the block as shown in the figure.


Solution

Let mass m and the rope of length l above the block be the system.

Mass of rope of length $l = \lambda l$

For the system under consideration, $(m + \lambda l)g - {T_P} = (m + \lambda l)a$

$ \Rightarrow {T_P} = (m + \lambda l)(g - a)$

Popular posts from this blog

A man starts walking from the point P (-3, 4) ....

A man starts walking from the point P (-3, 4), touches the x-axis at R, and then turns to reach at the point Q (0, 2). The man is walking at a constant speed. If the man reaches the point Q in the minimum time, then $50 [(PR)^2 + (RQ)^2 ]$ is equal to _ _ _ _ . Solution For time to be minimum at constant speed, the directions must be symmetric. In other words, the angles made by PR and RQ with the vertical must be the same just like in the law of reflection in optics. $tan \theta = \frac {MP}{MR} = \frac {NQ}{NR} $ $\Rightarrow \frac {3-r}{4} = \frac {r}{2}$ $\Rightarrow r=1 $ So, $R \equiv ( - 1,0)$ Now, $50(PR^2+RQ^2)=50[(4+16)+(1+4)]=1250$