# Inclined Plane Problems

1. May 22, 2015

### zesi

1. The problem statement, all variables and given/known data

A block with a mass of 100kg is being pulled with a force P in and effort to move it down the ramp. The ramp makes a 20 degree with the horizontal ground. The coefficient of the friction between the ramp and the block is 0.7.

Find the frictional force and the horizontal force.

Assume g=10m/s^2
2. Relevant equations

Inclined plane equations

3. The attempt at a solution

Let
F(y) = vertical force due to gravity.

F(py) = vertical force due to applied force.

F(x) = horizontal force due to gravity

F(px) = horizontal force due to applied force P.

F(yNet)
= Total vertical net force
= F(y) - F(py)
= mg(cos(theta)) - P(sin(theta))
= 1000cos20 - Psin20

F(xNet)
= Total horizontal net force
= F(x) + F(py)
= mg(sin(theta)) - P(cos(theta))
= 1000sin20 - Pcos20

Am I doing it right above?

Let
F(N) be the normal force.

My rationale is that since the horizontal force P lifts the mass upwards, it's force acting downwards will be reduced.

Am I right to say that,
F(N) = F(yNet)
Normal force = The net force acting vertically downware to the ramp.

Am I getting the concept right or wrong?

I am now stuck at this part because I have so many unknown. I still can't find P. Can someone correct my understanding? Thank you all.

2. May 22, 2015

### Orodruin

Staff Emeritus
What happened to the friction?

3. May 22, 2015

### zesi

Sorry let me re edit again for F(xNet)

Let F(f) = Frictional Force

F(xNet)
= Total horizontal net force
= F(x) + F(px) - F(f)
= mg(sin(theta)) + P(cos(theta)) - F(f)
= 1000sin20 + Pcos20 - F(f)

I still can't figure how to find the frictional force.

I know
F(f)
= F(N) * mu
= F(yNet) * 0.7
= 700cos 20 - 0.7Psin20

Did I make anymore mistakes? How do I move on from here?

Last edited: May 22, 2015
4. May 22, 2015

### collinsmark

I haven't gone through all your work. Your use of the word "net" is what is confusing me.

Assume for this problem that the force P is just large enough to get the block moving, but no more than that. If you prefer, you may think of the block sitting at rest on the ramp, but just almost enough to start moving. Or alternately, if it is moving, it is moving down the ramp at a constant velocity. My point is that it is not accelerating.

Given the lack of acceleration, what does Newton's second law of motion tell you about the net force on the block?

The next step is to break of the force equations into their x- and y- components. That will give you two simultaneous equations (one for the horizontal and another for the vertical). Two equations, two unknowns.

5. May 22, 2015

### zesi

I have broken down the forces and I am not sure if I am right. I will re-edit my working again.

(Note: Force P is parallel to the horizontal ground.)

F(gx) = Force parallel to the inclined due to gravity.

F(gy) = Force perpedicular to the inclined due to gravity.

F(px) = Force parallel to the inclined due to Force P.

F(py) = Force perpedicular to the inclined due to Force P.

Using Newton's second law, I will attempt to find the value P that allows it to move from rest. Net force = 0N

F(N) = The normal force perpendicular to the inclined.

F(f) = Frictional Force

For net force parallel to the ramp = 0N,
F(f) = F(gx) + F(px)
F(N)*mu = 1000sin20 + Pcos20
0.7(F(N)) = 1000sin20 + Pcos20
[Equation 1]

For net force perpendicular to the ramp = 0N,

F(N) = F(gy) - F(py)
F(N) = 1000cos20 - Psin20
[Equation 2]

Substitute equation 2 into equation 1.
0.7(1000cos20-Psin20) = 1000sin20+Pcos20

Eventually I get
P=267.80N approximately.

F(f)= F(N) * mu
F(f)
= 1000cos20-267.80sin20
= 848.10N

I get Force P=267.80N
Frictional Force = 848.10N

Both answers are still wrong. What am I doing wrongly?

Last edited: May 22, 2015
6. May 23, 2015