Question

Three point particles are fixed in position in an xy plane. Two of them, particle A of mass 5 g and particle B of mass 11 g, are shown in the figure with a separation of dAB = 0.461 m at angle θ = 30% Particle C, with mass 7 g, is not shown. The net gravitational force acting on particle A due to particles B and C is 2.74 x 1014 N at an angle of -163.8°. from the positive x axis. What are (a) the x coordinate and (b) the y coordinate of particle C? AB (a) Number: Units: (b) Number: Units:

Answer 1

The gravitational force acting on the particle A due to particle B is given by, 7. - Gimmie (cos 300(-i) + sin 30 (1) (667x10-" N-m/kg?) se(10% ke 10 kg 118 19 20 1g (cos 30º(-i)+sin 30º(j)) (0.461 m) = (1.49x10-44 N)(-i)+(8.63x10-5 N)(1) The gravitational force acting on the particle A due to particle C is given by, Ēic =(F_) (i)+(F;)(i) The net gravitational force acting on the particle A due to particles B and C is given by, Fax = (2.74x10*** N)(cos(-163.8°)(i)+sin (–163.8°) () = (2.63x10-14 N)(-i)+(7.64 x 10-15 )(-i) The net gravitational force acting on the particle A due to other two particles is given by, F.x = (1.49x10-24 N)(-i)+(8.63x10-15 )()+(F:)(i)+(F.)(3) (2.63 x 10-14 N)(-i)+(7.64x10-4 N)(-)=(-1.49x10-14 N+F)(f)+(8.63x10-4 N+F;)) From the above equation, -1.49x10-14 N+F, = -2.63x10-14 N F =-1.14x10-14 N 8.63x10-15 N+F, = -7.64 x10-15 N F, =-16.27x10-15 N

The angle made by the force due to particle C on particle A is given by, = tan an 11-16.27x10-15N) -1.14x10-14 N) = 54.98° The force due to particle C on particle A makes an angle of 54.98° with the negative x- axis in counterclockwise direction. The force due to particle Con particle A makes an angle of 180° +0 with the positive x- axis in counterclockwise direction. The horizontal component of the force is given by, F = F.e cos(180° +) -1.14x10-14 N=F. COS (180° +54.98°) -1.14x10-14 N cos(180° +54.989) = 1.99x10-14 N The magnitude of the gravitational force on the particle A due to particle C is given by, Gm mc FAC - 1 dic ) (6.67x10-11 N.mp/kg (10-kg . 10-kg 1 golg 1.99x10-14 N = (10-kg (6.67x10-5 N·mº / kg*) 58(102 (1.99x10-14 N) 10-9 kg 1 g 19 = 0.3425 m

The x coordinate of the particle C is given by, x=dA cos (180° +) =(0.3425 m) cos(180° +54.98) =-0.196 m

(b) The y coordinate of the particle C is given by, y=d.csin (180°+0) =(0.3425 m sin(180° +54.98) = -0.280 m