Solving Inclined Plane Problems

AI Thread Summary
The discussion revolves around solving a physics problem involving a block on an inclined plane, where a 100kg block is pulled down a ramp at a 20-degree angle with a friction coefficient of 0.7. Participants analyze the forces acting on the block, including gravitational, applied, and frictional forces, while attempting to derive equations for net forces in both vertical and horizontal components. The user struggles to find the correct value for the pulling force P and the frictional force, leading to confusion about the application of Newton's second law and the role of the normal force. Ultimately, the user arrives at approximate values for P and the frictional force but doubts their accuracy, seeking clarification on their calculations and understanding of the concepts involved.
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Homework Statement



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.

ppSDyCA.jpg


Assume g=10m/s^2

Homework Equations



Inclined plane equations

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.
 
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What happened to the friction?
 
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:
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. :wink:
 
I have broken down the forces and I am not sure if I am right. I will re-edit my working again.

CVvO6ZI.jpg


(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:
Something goes wrong during your 'Eventually'. Please post the detailed working.
 
Kindly see the attached pdf. My attempt to solve it, is in it. I'm wondering if my solution is right. My idea is this: At any point of time, the ball may be assumed to be at an incline which is at an angle of θ(kindly see both the pics in the pdf file). The value of θ will continuously change and so will the value of friction. I'm not able to figure out, why my solution is wrong, if it is wrong .
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