Answer:
weight at height = 100 N .
Explanation:
The problem relates to variation of weight due to change in height .
Let g₀ and g₁ be acceleration due to gravity , m is mass of the object .
At the surface :
Applying Newton's law of gravitation
mg₀ = G Mm / R²
At height h from centre
mg₁ = G Mm /h²
Given mg₀ = 400 N
400 = G Mm / R²
400 = G Mm / (6400 x 10³ )²
G Mm = 400 x (6400 x 10³ )²
At height h from centre
mg₁ = 400 x (6400 x 10³ )²/ ( 2 x 6400 x 10³)²
= 400 / 4
= 100 N .
weight at height = 100 N
Let F1=Force exerted by the brother (+F1)
F1= Force exerted by the sister (-F2)
Fnet=(+F1) + (-F2)
Fnet= (+F1) + (-F2)
Fnet=F1 - F2
Fnet= (+3N)+(-5N)
Fnet= -2N
-F
towards the sister (-F) (greater force applied)
Answer:
.409 N
Explanation:
For this to balance, the moments around the fulcrum must sum to zero.
On the left you have .21 ( is that down? I will assume it is)
Counterclockwise moments :
.21 * 40 + 1.0 * 20
Clockwise moments :
.5 * 20 + F * 45
these moments must equal each other
.21*40 + 1 *20 = .5 * 20 + F * 45
F = .409 N
