how to get the force an object is on a lever?

Question

you have a 100 lb on one side and 10 lb on the other and the boards fulcrum point is 40'' on one side and 60'' on the other how much force will the 10 lb have aginst it

the 100 lb is on the 40'' side and the 10'' lb is on the 60'' side what is the thrust on the 10 lb (going up )?

Answer 1

Interesting sea-saw problem.

40 x 100 = 60 x 10 + 60 x F
so the trust F you looking for is
F= [40 x 100 - 60 x 10 ]/60
F=[400 - 60]/6=56.7 lb

Answer 2

I can't quite understand what is going on in this problem. You have a lever and some weights. What do you mean by the question

how much force will the 10 lb have aginst it?

Do you want to know the net force at the point of the 10lb force?

This isn't well worded at all.

Answer 3

m1 = 100 lb
m2 = 10 lb
r1 = 40"
r2 = 60"

Force exerted by the lever on heavy block F1
Force exerted by the lever on light block F2

Since lever is weightless, net momentum acting on the
lever must be zero:
- F1 r1 + F2 r2 = 0
F1 r1 = F2 r2
Not surprisingly the equation above is equivalent to
Archimedes principle.

Angular velocity of the lever is ω.
Velocity of heavy block is v1 = - ωr1
Velocity of light block is v2 = + ωr2

Acceleration of heavy block is a1 = dv1/dt = -r1 dω/dt
Acceleration of light block is a2 = dv2/dt = +r2 dω/dt

Second law of Newton
m1a1 = -gm1 + F1
m2a2 = -gm2 + F2

a1 = -g + F1/m1
a2 = -g + F2/m2

-r1 dω/dt = -g + F1/m1
+r2 dω/dt = -g + F2/m2

-r1 r2 dω/dt = r2(-g + F1/m1)
+r2 r1 dω/dt =r1(-g + F2/m2)

(r1 + r2)g = r2 F1 / m1 + r1 F2 / m2

Substitude F1 = r2/r1 F2:
r2(r1 + r2)g = (r2²/m1 + r1²/m2) F2

Answer:
F = r2 m1 m2 (r1 + r2) g / (m2r2² + m1r1²)