I understand now, I kept thinking N was Newtons times μ instead of the Normal force. The normal force is equal and opposite the downward forces Fp and MGp hence those to added together equal it.
Thanks
Okay so the acceleration of the barrel is a = Fh/m = cos(50°)(320N)/24 = 8.57m/s^2
Putting that into the equation
V^2=Vi^2 + 2a(Δx)
(Δx) = V^2-Vi^2/(2a)
(Δx) = 0-(1.2^2)/(2*8.57)
= 0.084 meters
Yes the Normal for going up perpendicular to the ramp has a magnitude equal to the sum of the downward forces Fp and MGp right?
So applying that to the forces parallel to the ramp:
Fh = Nμ + MGh
cos(50°)(320N) = N(.48) + mg(sin(50°)
N= 53.163
So the sum of the forces up and down the plane equal Fh-Fμ-MGh which is:
Fμ = Fh*μ where μ is the kinetic friction constant to .48?
If the above is correct then:
cos(50°)(320N) - ((.48)cos(50°)(320N))- (sin(50°)(24kg)(9.8)
= 205 - 98.7 - 180 which would be negative value for the force...
μ
Okay I think these are the six forces after resolving the 320N force into x and y components and the force mg into x and y components.
Fh= cos(σ)(320N)
Fp= sin(σ)(320N)
MGh= sin(σ)mg
MGp= cos(σ)mg
EDIT:
So the sum of the forces up and down the plane equal Fh-Fμ-MGh?
Diagram now has 6 forces.
Fh is the force parallel to the ramp/plane which is pulling the barrel up
Fp is the force perpendicular to the ramp/plane. This is equal and opposite to the Normal force correct?
Fh= cos(50°)320N
Fp= cos(50°)mg
How to solve for acceleration?
Ah, right friction force is opposite direction of motion.
The extra x in that equation was a mistake, thanks.
So First I solved for the force going up the ramp which would be cos(σ) component of the horizontal force of 320N. Then solving for acceleration since the mass is given...
Homework Statement
Problem:
"On the 50 degree ramp below an empty barrel is sliding up it to get filled at a factory. The empty barrel has a mass of 24kg and has an initial velocity of 1.2m/s. The kinetic coefficient of friction between the barrel and the ramp is μ =0.48 Finally a...