Six objects are placed at the vertices of a r | vedclass

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

Question

(A) For a mass $m$ at the centroid of the hexagon (at the origin), the net force is zero when $\Sigma F_x = 0$ and $\Sigma F_y = 0$.The gravitational

Vedclass · Accepted Answer

$0$

Vedclass · Answer

(A) For a mass $m$ at the centroid of the hexagon (at the origin), the net force is zero when $\Sigma F_x = 0$ and $\Sigma F_y = 0$.The gravitational force exerted by a mass $m_k$ at distance $r$ from the origin on mass $m$ is $F_k = \frac{G m m_k}{r^2}$.The $x$-component of the force is $F_{kx} = -\frac{G m m_k}{r^2} \cos 	heta_k$ (directed towards the mass).Summing the $x$-components: $\Sigma F_x = -\frac{G m M}{r^2} \sum_{k=1}^{6} k^i |\cos 	heta_k| \cos 	heta_k = 0$.For the vertices of a regular hexagon, the angles $	heta_k$ are $0^{\circ}, 60^{\circ}, 120^{\circ}, 180^{\circ}, 240^{\circ}, 300^{\circ}$.Substituting these values:$1^i |\cos 0^{\circ}| \cos 0^{\circ} + 2^i |\cos 60^{\circ}| \cos 60^{\circ} + 3^i |\cos 120^{\circ}| \cos 120^{\circ} + 4^i |\cos 180^{\circ}| \cos 180^{\circ} + 5^i |\cos 240^{\circ}| \cos 240^{\circ} + 6^i |\cos 300^{\circ}| \cos 300^{\circ} = 0$$1^i(1)(1) + 2^i(1/2)(1/2) + 3^i(1/2)(-1/2) + 4^i(1)(-1) + 5^i(1/2)(-1/2) + 6^i(1/2)(1/2) = 0$$1^i + \frac{2^i}{4} - \frac{3^i}{4} - 4^i - \frac{5^i}{4} + \frac{6^i}{4} = 0$Testing $i=0$:$1 + \frac{1}{4} - \frac{1}{4} - 1 - \frac{1}{4} + \frac{1}{4} = 0$. This holds true.Thus, the value of $i$ is $0$.

Similar Questions

Gravitational mass is proportional to gravitational

The gravitational field in a region is given by $\vec{E} = (5\,N/kg)\,\hat{i} + (12\,N/kg)\,\hat{j}$. If the potential at the origin is taken to be zero,then the ratio of the potential at the points $(12\,m, 0)$ and $(0, 5\,m)$ is:

$A$ spherical shell is divided into two parts. If the gravitational intensity at point $P$ due to the upper part is ${I_1}$ and due to the lower part is ${I_2}$,then:

The gravitational field,due to the 'left over part' of a uniform sphere (from which a part as shown,has been 'removed out'),at a very far off point,$P$,located as shown,would be (nearly)

The figure shows a hemispherical shell. The direction of the gravitational field intensity at point $p$ will be along

Vedclass Test Series

Exam Paper Generator

Online Exam Module