Question

A rectangular plank of mass "M" and length "L" is placed along the x axis. Another mass "m" is kept at a distance of "a" from one end of the plank on the same axis. Calculate the net gravitational force acting between the objects.



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Answer 1

Given :

Mass of rectangular plank = M

Length of rectangular plank = L

It is placed along x axis.

Mass m is kept at a distance of a from one end of the plank on the same axis.

To Find :

Net gravitational force acting between the objects.

Solution :

❖ The magnitude of gravitational force between two point masses is directly proportional to the product of masses and inversely proportional to the square of the distance between them.

• This is known as newton's law of gravitation.

Mathematically, $\underline{\boxed{\bf{\orange{F=\dfrac{Gm_1m_2}{r^2}}}}}$

Where r is the distance between two point masses.

• In this question we can't directly put values in the formula as separation between two object is smaller in comparison to masses.

Let's consider a small element of mass dm and length dr. Let distance between this small element and mass m be r.

★ Linear mass density of rectangular plank is given by

• λ = M / L

Mass of small element in terms of linear mass density will be

• dm = λ dr = (M / L) dr

Gravitational force acting between masses m and dm can be calculated by using general formula.

$\sf:\implies\:dF=\dfrac{Gm(dm)}{r^2}$

In order to find net gravitational force acting between rectangular plank and mass m, we have to integrate the above equation from a to (L + a).

$\displaystyle\sf:\implies\:F=\int\limits_a^{L+a}\:dF=\int\limits_a^{L+a}\:\dfrac{Gm(dm)}{r^2}$

$\displaystyle\sf:\implies\:F=\int\limits_a^{L+a}\:\dfrac{Gm}{r^2}\cdot\dfrac{M}{L}dr$

$\displaystyle\sf:\implies\:F=\dfrac{GMm}{L}\int\limits_a^{L+a}\:\left(\dfrac{1}{r^2}\right)\:dr$

$\sf:\implies\:F=\dfrac{GMm}{L}\left[-\dfrac{1}{r}\right]_a^{L+a}$

$\sf:\implies\:F=\dfrac{GMm}{L}\left[-\dfrac{1}{(L+a)}-\left(-\dfrac{1}{a}\right)\right]$

$\sf:\implies\:F=\dfrac{GMm}{L}\left[\dfrac{1}{a}-\dfrac{1}{L+a}\right]$

$\sf:\implies\:F=\dfrac{GMm}{L}\left[\dfrac{L+a-a}{a(L+a)}\right]$

$\sf:\implies\:F=\dfrac{GMm}{L}\times\dfrac{L}{a(L+a)}$

$\bf:\implies\:F=\dfrac{GMm}{a(L+a)}$

If a >> L then;

$:\implies\:\underline{\boxed{\bf{\gray{F=\dfrac{GMm}{a^2}}}}}$

Answer 2

L/4

Let distance moved by the planck be x, then wrt ground the man will move by a distance L−x

Now the net displacement of the centre of mass will be zero in ground frame

M(L−x)=M∗x/3

x=3∗L/4

(L−x)=L/4