Motion on Inclined ramp with Friction Problem

[DividedbyZero]

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 horizontal force is acting on the barrel of 320N.
1.)What is the barrel's acceleration?
2.)What is the direction of the barrel's acceleration?
3.) How far up the ramp does the barrel go before its velocity is zero?"

First picture is of what is given

Homework Equations

F=ma (of course)
accelerations will involve x or y components with regard to theta (sin(σ) and cos(σ))
?Kinetic formulas? Δx=

The Attempt at a Solution

Second picture attachment is a picture of my diagram so far. Is it missing a force going down the ramp? Parallel to the friction force?

A.)Not sure where to start here.

B.) The acceleration is down the ramp because it is slowing to a stop. Correct?

C.) Solvable after I find acceleration in part A? Using Δx= Vi(t)+ 0.5(a)t^2 where Vi is initial velocity. Or on second thought perhaps this one if you can not find time: V^2=Vi^2 + 2a(xΔx)

Please and thank you!

 

[azizlwl]

In your second diagram the direction of the frictional force should be in opposite way, pointing downward since it is sliding up.

Then you can resolve(replace) the 4 forces in only direction of parallel and tangent to plane.
From this diagram the net tangent forces to the plane is zero.
The net parallel forces will accelerate the barrrel up or down.

 

[PhanthomJay]

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 horizontal force is acting on the barrel of 320N.
1.)What is the barrel's acceleration?
2.)What is the direction of the barrel's acceleration?
3.) How far up the ramp does the barrel go before its velocity is zero?"

First picture is of what is given

Homework Equations

F=ma (of course)
accelerations will involve x or y components with regard to theta (sin(σ) and cos(σ))
?Kinetic formulas? Δx=

The Attempt at a Solution

Second picture attachment is a picture of my diagram so far. Is it missing a force going down the ramp? Parallel to the friction force?

You've identified all the forces acting; Now you have to break them up into their vector components parallel and perpendicular to the plane, then use the applicabe laws of Newton in the y and x direction separately to sove for N and a.

A.)Not sure where to start here.

B.) The acceleration is down the ramp because it is slowing to a stop. Correct?

As per question 3, this is apparently correct, although you'd have to crunch out the numbers to solve for the correct direction of a.

C.) Solvable after I find acceleration in part A? Using Δx= Vi(t)+ 0.5(a)t^2 where Vi is initial velocity. Or on second thought perhaps this one if you can not find time: V^2=Vi^2 + 2a(xΔx)

this last one is good, but you have an extra x term in there. Don't get too fancy doing calculus when you don't need to.
And don't forget to ask yourself what is going on in the y direction perpendicular to the plane.

 

[DividedbyZero]

In your second diagram the direction of the frictional force should be in opposite way, pointing downward since it is sliding up.

Then you can resolve(replace) the 4 forces in only direction of parallel and tangent to plane.
From this diagram the net tangent forces to the plane is zero.
The net parallel forces will accelerate the barrrel up or down.

Ah, right friction force is opposite direction of motion.

You've identified all the forces acting; Now you have to break them up into their vector components parallel and perpendicular to the plane, then use the applicabe laws of Newton in the y and x direction separately to sove for N and a. As per question 3, this is apparently correct, although you'd have to crunch out the numbers to solve for the correct direction of a. this last one is good, but you have an extra x term in there. Don't get too fancy doing calculus when you don't need to.
And don't forget to ask yourself what is going on in the y direction perpendicular to the plane.

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:

cos(50°)(320N) = 205.69N
205.69N = ma
a= 205.69/24kg = 8.53m/s^2

I feel like this is the force if it was stationary or something. How do I account for the frictional force?

The below?
μ= kinetic friction constant

So if the total force up the ramp would be the force up the ramp minus the frictional force pulling it down the ramp that would be 205.69-(205.69*μ) = 106.96N

Then I would have to recalculate the acceleration since 106.96N is the 'new' non-zero net force up the ramp. 106.96N/24kg= 4.45m/s^2

Are either of these correct? If 4.45m/s^2 is the correct acceleration of the barrel then

V^2=Vi^2 + 2a(Δx)
(Δx) = V^2-Vi^2/(2a)
(Δx) = 0-(1.2^2)/(2*4.45) →(this equals negative but assuming negative accel. to come to a stop?)
= .1617 meters with sigfigs: 0.16m

What do you think?

 

[azizlwl]

Now you remove the mg and replace this force with parallel to plane and another perpendicular or resolving mg to resultant of 2 forces. Now you have 2 forces downward parallel to plane, component of mg and friction.

Likewise you have to replace the 320N force to 2 forces. Parallel and perpendicular to plane.

You will see that friction depends on N and N is result of 2 component forces.

 

[DividedbyZero]

Now you remove the mg and replace this force with parallel to plane and another perpendicular or resolving mg to resultant of 2 forces. Now you have 2 forces downward parallel to plane, component of mg and friction.

Likewise you have to replace the 320N force to 2 forces. Parallel and perpendicular to plane.

You will see that friction depends on N and N is result of 2 component forces.

So you are saying resolve the force going up the ramp into two forces; one parallel to the horizontal and one perpendicular to the the first force?

 

[azizlwl]

Use your previous diagram with the mg pointing downward and 320N horizontally pointing to the right. The normal force and friction are in correct position.

The only thing to do is erase the mg and replace with 2 forces, perpendicular and horizontal to plane.
Equally do it to 320N force.

What you should get 6 forces acting on the barrel.
2 downward, 1 upward parallel to plane
2 downward, 1 upward perpendicular to plane.

add: fbd with 6 forces.

 

[DividedbyZero]

Use your previous diagram with the mg pointing downward and 320N horizontally pointing to the right. The normal force and friction are in correct position.

The only thing to do is erase the mg and replace with 2 forces, perpendicular and horizontal to plane.
Equally do it to 320N force.

What you should get 6 forces acting on the barrel.
2 downward, 1 upward parallel to plane
2 downward, 1 upward perpendicular to plane.

add: fbd with 6 forces.

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?

 

[azizlwl]

Once you resolve a force, you remove it from the diagram. It does not exist by itself anymore.

You've done correctly resolving to parallel of the plane.(320N = component #1 + component #2)
You should do resolve perpendicular to the plane.
Means you're replace 320N to 2 components.

Do the same with mg.


Edit: Sorry ,You're resolve F_p. But it has nothing to do with mg.
Good you have done Fh,Fp(wrong calculation) and now add MGh and MGp

 

[DividedbyZero]

μ

Once you resolve a force, you remove it from the diagram. It does not exist by itself anymore.

You've done correctly resolving to parallel of the plane.(320N = component #1 + component #2)
You should do resolve perpendicular to the plane.
Means you're replace 320N to 2 components.

Do the same with mg.


Edit: Sorry ,You're resolve F_p. But it has nothing to do with mg.
Good you have done Fh,Fp(wrong calculation) and now add MGh and MGp

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?

 

[azizlwl]

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?

That's correct fbd.
Friction should be μN.

 

[DividedbyZero]

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. Does this make since?
-73.7N=24kg(a) solve for acceleration here?

a=-3.07 m/s^2

Negative acceleration because the barrel is slowing down?

 

[azizlwl]

There's error in your calculation.
Frictional force=μN NOT Fμ or Fhμ

Find the value of N.

 

[DividedbyZero]

There's error in your calculation.
Frictional force=μN NOT Fμ or Fhμ

Find the value of N.

I don't have the frictional force to solve for N do I? Unless its MGp the force down on the barrel?

 

[azizlwl]

You see from the diagram there are 2 forces pulling/pressing the object down.
By Newton 3'rd law there must be equal opposite force, where we call normal force.

Or you see that the barrel no going down or up perpendicullaly , net force equal to zero.

 

[DividedbyZero]

You see from the diagram there are 2 forces pulling/pressing the object down.
By Newton 3'rd law there must be equal opposite force, where we call normal force.

Or you see that the barrel no going down or up perpendicullaly , net force equal to zero.

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

 

[azizlwl]

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? 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

You have right statement. From this statement you should know the value of N.

 

[DividedbyZero]

You have right statement. From this statement you should know the value of N.

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

 

[azizlwl]

Maybe you can show the net horizontal force along the plane in term of MGh, MGp, Fh,Fp, and μ variables. Refer to your fbd.

 

[DividedbyZero]

Maybe you can show the net horizontal force along the plane in term of MGh, MGp, Fh,Fp,N and μ variables. Refer to your fbd.

Nμ + MGh + Fh ... or so I thought

 

[azizlwl]

F_net=Fh-Nμ-MGh
N=MGp+Fp
F_net=Fh-μ(MGp+Fp)-MGh

I hope you understand why i replace N with MGp+Fp.

You have done good at resolving the forces.
I think you have to know more about N or normal force and frictional force.
And the rest refer to PhantomJay.

 

[DividedbyZero]

F_net=Fh-Nμ-MGh
N=MGp+Fp
F_net=Fh-μ(MGp+Fp)-MGh

I hope you understand why i replace N with MGp+Fp.

You have done good at resolving the forces.
I think you have to know more about N or normal force and frictional force.
And the rest refer to PhantomJay.

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