First slide

Gravitational force and laws

Question

ABC is a right angled isosceles triangle with AB = BC and $B = 90^{0}$ . Two body of mass ‘m’ are placed at A and C, the resultant force on 2kg mass placed at ‘B’ make an angle ….with line BC is

Moderate

A

$60^{0}$
B

$45^{0}$
C

$90^{0}$
D

$30^{0}$

Solution

Force on 2kg due to mass ‘m’ at A and B is $F_{1} = F_{2}$

$∴$ Resultant force makes an angle $45^{0}$ with ‘BC’
$u n i v e r s a l g r a v i t a t i o n c o n s \tan t = G; m a s s = m; d i s \tan c e = r; F o r c e o f a t t r a c t i o n b e t w e e n t w o m a s s e s = \frac{G m_{1} m_{2}}{r^{2}} f r o m p a r a l l e \log r a m l a w o f a d d i t i o n o f v e c t o r s, r e s u l \tan t f o r c e m a g n i t u d e i s F = \sqrt{{F_{1}}^{2} + {F_{2}}^{2} + 2 F_{1} F_{2} \cos θ} h e r e θ = 90 \Rightarrow \cos 90 = 0 h e n c e F_{R} = \sqrt{{F_{1}}^{2} + {F_{2}}^{2}} \sin c e F_{1} = F_{2} = F F_{R} = \sqrt{{F_{1}}^{2} + {F_{1}}^{2}} F_{R} = \sqrt{2} F_{1} \tan θ = \frac{F_{2} S i n 90}{F_{1} + F_{2} C o s 90} = 1 = \tan 45 θ = 45$

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