Question

a.What is the magnitude of the net gravitational force on the m1=21kg mass in the figure?(Figure 1) Assume m2=m3=10kg.

Express your answer using two significant figures.

b.What is the direction of the net gravitational force on the m1=21kg mass in the figure? Assume m2=m3=10kg.

Having problems loading image. You can look up image in google image the first image m2 =5.0 cm, m3=5.0 cm, Height is 20.0 cm

Answer 1

Distance between mass 1 and mass 2

D^2 = 20^2 + 5^2

D^2 =425cm^2


Gravitation force = G m1 m2 / D^2

G = 6.67 x 10^-11

= 6.67 x 10^-11 ( 10kg)(21kg) /0.0425 m^2
= 3.2956 X 10^-7N
Direction for mass 2 = 90 + invtan (5/20)= 104.03 degrees

Force for mass m1,m3 combination
Force is the same =3.2956 X 10^-7N

direction = 90 - invtan (5/20) = 75.96 degrees

resultant is the vector sum
x component m1,m2 combination = 3.2956 X 10^-9N cos (104..03) = -7.989 x10^-8
x components m1,m3 combination = 3.2956 X 10^-9N cos (75.96) = 7.995 X10^-8

y component m1,m2 combination = 3.2956 X 10^-9N sin (104..03)= 3.1972 X 10^-7
y components m1,m3 combination = 3.2956 X 10^-9N sin (75.96)= 3.1972 x 10^-7

The resultant is the addition of the x and y components

the x components net to zero
the y components v add to 6.4x 10^-7......ANSWER

since the x component cancel out the resultant is along the y axis
angle = 90 degrees ANSWER
Hope this helps you