Question

Four masses are positioned at the corners of a rectangle, as indicated in the figure.

(a) Find the magnitude and direction of the net force acting on the 2.0-kg mass
(b) Find the magnitude and direction of the net force acting on the 1.0-kg mass
(c) Find the magnitude and direction of the net force acting on the 3.0-kg mass
(d) Find the magnitude and direction of the net force acting on the 4.0-kg mass

1.0 kg 0.20 m 2.0 kg 0.10 m ! 4.0 kg 3.0 kg

Answer 1

1.0 kg 1 0.20 m 12 ----- F + 26.569 2.0 kg 26.5662 F24 F23 - Find \\F13 0.10 m ! F41 F42 F31 F32 26.560 26,56 F43 4.0 kg F34 3.0 kg

a)

Force on
m2 = 2.0 kg
due to other 3 masses ,
F2 = F21 + F23 + F24

1.0 Gmim2 Gm2m4 cos 26.56° F21 = -F21,1 – F24,1 = - -= -6.67 * 10-11 * 2.0 4.0* (10.05) = -1.29 * 10-8 N r12 TRA 0.202

Gm3m2 Gm2m4 sin 26.56° F2y = -F23,y - F24,y = – 3.0 4.0 * 1 = -6.67 * 10-11 * 2.0 = -4.48 * 10-8 N RŽA 0.102 + 10.052
Magnitude of net force on $m_2$ is  
F2= V1.292 + 4.482 * 10-8 = 4.46 * 10-8 N

0 = tan-1-4.48 -1.29 6 = 73.90

Direction is   counter clockwise from +X axis.

180 +9=180 + 73.9= 253.9"

----------------------

b)

Force on
m1 = 1.0 kg
due to other 3 masses ,
F1 = F12 + F13 + F14

2.0 Gmim2 Gm¡m3 cos 26.56° F1 = +F12,1 + F13,1 = - * = 6.67 * 10-11 * 1.0 ris 3.0 * * (0.05) = 6.91 * 10-9N 712 0.202 +

Gmım4 Gm¡m3 sin 26.56° 4.0 3.0* Fly = -F14,y – F13, y = = -6.67 * 10-11* 1.0 = -2.85 * 10-8 N ría 0.102 10.052
Magnitude of net force on $m_1$ is  
F1 = V0.6912 + 2.852 * 10-8 = 2.93 * 10-8 N

A = tan-1 -2.85 0.69176.40

Direction is    counter clockwise from +X axis

3Ꮾ0 - Ꮎ = 360 -76.4 = 283.6

--------------------.

c)

Force on
m3 = 3.0 kg
due to other 3 masses ,
F3 = F31 + F32 + F34

4.0 Gm3m4 Gm¡m3 cos 26.56° 1.0 * 2 F31 = -F31,1 – F34,1 = - -= -6.67 * 10-11 * 3.01 - 0.202' (0.05) = -2.36 * 10-8 N ris

Gm3m2 Gmzmı sin 26.56° F3y = F32, y + F31, y = = . = 6.67 * 10-11 * 3.0 2.0 0.102 1.0 * A (0.05) = 4.18 * 10-8N Th3
Magnitude of net force on $m_3$ is  
F3 = V2.362 + 4.182 * 10-8 = 4.80 * 10-8 N

0 = tan-1 4.18 -2.366 0.6

Direction is   counter clockwise from +X axis.

180 -9=180 - 60.6 = 119.4°

----------------------

d)

Force on
m4 = 4.0 kg
due to other 3 masses ,
F4 = F41 + F12 + F43

2.0 * Gm4m3 Gm4m2 cos 26.56° F41 = F43,1 + F42,1 = = -= 6.67 * 10-11 * 4.0 + = 2.96 * 10-8 N 0.202 + (0.05)

Gmami Gm4m2 sin 26.56° F4y = F41,y + F42,y = - . = 6.67 * 10-11 * 4.0 1.0 2.0 * A 0.102" (0.05) = 3.14 * 10-8 N 141
Magnitude of net force on $m_4$ is  
F4 = 12.962 + 3.142 * 10-8 = 4.32 * 10-8 N

0 = tan-1 2.96 = 46.89

Direction is  $46.8\degree$ counter clockwise from +X axis.

---------------