Question

4 The figure below shows three masses

What are the magmtude and the direction of the net gravitational force on the 20 0 kg mass? Give the direction as an angle

Answer 1

Given : Three masses of 10kg, 10kg and 20kg respectively, are located in X - Y plane as shown in figure.

To find : The net gravitational force on the 20kg mass. and also give the direction of forces.

solution : if we join the masses, an isosceles triangle ABC will be formed as shown in diagram.

using cosine formula from properties of triangle

cosC = (a² + b² - c²)/2ab

here, a = BC = √(5² + 20²) = √425

b = AC = √(5² + 20²) = √425

c = AB = 5 + 5 = 10cm

now, cosC = (425 + 425 - 100)/2 × 425

= (750)/850

= 15/17 .......(1)

force between A and C, F₁ = GmM/r²

here, m = 10kg, M = 20kg, r = √425

so, F₁ = (6.67 × 10¯¹¹ × 10 × 20)/(425)

= 3.14 × 10¯¹¹ N

similarly, force between B and C, F₂ = GmM/r² = 3.14 × 10¯¹¹ N

now angle between F₁ and F₂ is C

so, Fnet = √{F₁² + F₂² + 2F₁.F₂cosC}

= √{2(3.14 × 10¯¹¹)² + 2(3.14 × 10¯¹¹)² × 15/17}

= (3.14 × 10¯¹¹) √(2 + 1.76)

= 3.14 × 1.939 × 10¯¹¹

= 6.088 × 10¯¹¹ N

Therefore net gravitational force on the 20kg mass is 6.088 × 10¯¹¹ N

angle made by resultant with gravitational force, F₁ , Φ = tan¯¹ [F₂sinC/(F₁ + F₂cosC)]

= tan¯¹[ (3.14 × 10¯¹¹) × 8/17/(3.14 × 10¯¹¹ + 3.14 × 10¯¹¹ × 15/17]

= tan¯¹ [(8/17)/(1 + 15/17)]

= tan¯¹ [ 8/32 ] = tan¯¹ [1/4]