Three particles each of mass M are placed at | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(B) Let the equilateral triangle be $PQR$ with side length $l$. The masses $M$ are at $P, Q,$ and $R$. The mass $m$ is placed at the midpoint $L$ of s

Vedclass · Accepted Answer

$\frac{4GMm}{3l^2}$

Vedclass · Answer

(B) Let the equilateral triangle be $PQR$ with side length $l$. The masses $M$ are at $P, Q,$ and $R$. The mass $m$ is placed at the midpoint $L$ of side $QR$.The gravitational force exerted by the mass $M$ at $Q$ on mass $m$ at $L$ is $F_Q = \frac{GMm}{(l/2)^2}$ directed towards $Q$.The gravitational force exerted by the mass $M$ at $R$ on mass $m$ at $L$ is $F_R = \frac{GMm}{(l/2)^2}$ directed towards $R$.Since $F_Q$ and $F_R$ are equal in magnitude and opposite in direction,they cancel each other out $(F_Q + F_R = 0)$.The net force on $m$ is therefore only due to the mass $M$ at vertex $P$.The distance $PL$ is the altitude of the equilateral triangle,given by $PL = l \sin 60^{\circ} = l \frac{\sqrt{3}}{2}$.The magnitude of the force is $F = \frac{GMm}{(PL)^2} = \frac{GMm}{(l\sqrt{3}/2)^2} = \frac{GMm}{3l^2/4} = \frac{4GMm}{3l^2}$.

Three particles each of mass $M$ are placed at the three corners of an equilateral triangle of side $l$. What is the gravitational force due to this system of particles on another particle of mass $m$ placed at the midpoint of any side?

Similar Questions

Two spheres of mass $m$ and $M$ are situated in air and the gravitational force between them is $F$. The space around the masses is now filled with a liquid of specific gravity $3$. The gravitational force will now be

If two identical spherical bodies of same material and dimensions are kept in contact,the gravitational force between them is proportional to $R^{x}$,where $x$ is a non-zero integer. [Given: $R$ is the radius of each spherical body]

$A$ mass $M$ is split into two parts $m_0$ and $M-m_0$. These two masses are then separated by a distance $D$. If the gravitational force between the parts is maximum,then the ratio $\frac{m_0}{M}$ is

$A$ straight rod of length $L$ extends from $x = a$ to $x = L + a$. The gravitational force it exerts on a point mass $m$ at $x = 0$,if the mass per unit length of the rod is $A + Bx^2$,is given by

The distance of the planet Pluto from the Sun is $40$ times the distance of the Earth from the Sun. Assuming the masses of the Earth and Pluto are equal,find the ratio of the gravitational force exerted by the Sun on the Earth to that on Pluto.

Vedclass Test Series

Exam Paper Generator

Online Exam Module