How Does a Speck of Dust Affect the Motion of a Box on an Inclined Ramp?

In summary: Ma block? (In the Y direction)If I understand you correctly.. I should have something like this:Force Gravity on Block = (Ma)(9.8)sin(30) = Ma(4.9)NF = 0Ma(4.9) = 0Ma = 0Hmm, not quite sure what that means, would it be:(100gcos(30) - Tension2*sin(22))/Ma= AI'm not sure what you're asking. I don't know what you mean by "would it be".I'm sorry I'm not being clear, what I mean is:If I am
  • #1
joemost12
20
0

Homework Statement



100 kg box is held in place on a ramp that rises at 30° above the horizontal. There is a massless rope joint to the box that makes a 22° angle above the surface of the ramp. Coefficients of friction between the box and the surface of the ramp are μk = 0.40 and μs = .60. The pulley has a mass of 10 kg and a radius of 10 cm. However the pulley does not have a friction.

A). What is the Max weight the hanging mass can have so that the box remains at rest.

B). The system is at rest with the weight of the hanging mass found in Part A, but a speck of dust comes to rest on the weight on the hanging mass, making the system unstable. What is the acceleration of the box at the very moment it starts moving?See diagram
https://i.imgur.com/xCWIHWb.png
xCWIHWb.png


Homework Equations

The Attempt at a Solution



So I think I got part A, and sort of got part B setup.. but I am not really sure where to go from there. [/B]

Part A)
Here is how I setup my equations:
Force Gravity on Block = (100)(9.8)sin(30) = 490
NF = 100cos(30) - Tension*sin(22) = 50√3 - Tension*sin(22)
Mass on Ramp = Ma
Mb =
Hanging mass = Tension-(Mb)(9.8) = 0

Net force on Box = 0, forces equation =
Tension*cos(22) - Magsin(30) - ( Magcos(30) - Tension*sin(22) )μ = 0

I then proceeded to plug in Ma, g = 9.8, and μ = .60
Then using the equation above and the Hanging mass equation I solved for Mb

I got
Mb = ~ 88 kg

Part B)
I am fairly confident about my answer of Part A.. but now unsure exactly how to solve part B.
I know the radius of the pulley is 10 cm and the mass is 10 kg.


I had to find the equations for the two tensions. However, one thing I had to form a new equation for (at least I think ) is the Normal Force. Before I knew the vertical component of the normal force would just be 0, but now since it is moving it could also change.

My new equation is:
NF = 100cos(30) - Tension*sin(22) = Ma A
NF = 100cos(30) - Tension*sin(22) - Ma A = 0

I know I need to substitute this into my original equation, so I can use it to calculate the force of Friction
Tension2*cos(22) - Magsin(30) - ( 100gcos(30) - Tension2*sin(22) - Ma A )μ = 0
put this equation in terms of Tension2, but I must leave acceleration in terms of A
Tension2 = 129A + 770
My equation for tension1 was easier, just:
Tension1 = Mb A - Mb A

Now I plug them into a Torque equation taking into account the mass of the pulley and radius
Tnet = [ Tension2 - Tension1]*radius = .5 m r^2 ∝

Now here is where my struggles appear, I have two unknowns with A and Mb. I've got to imagine that the intention of Mass B is that the speck of dust would represent just about nothing, and the point of it was to explain that the system would be in motion and I should just account for it as the same mass from before, but I don't know. ANy advice on this would be great!Let me know if there is anything I should elaborate on something.. ,thanks in advance!

 
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  • #2
WELCOME TO PF!
Part (A) looks correct.
joemost12 said:
Part B)
My new equation is:
NF = 100cos(30) - Tension*sin(22) = Ma A
This equation says that NF = MaA which implies that the net force accelerating Ma is the normal force. Is that correct?
Also, the middle expression in your equation is missing a factor of g somewhere.

Tension2*cos(22) - Magsin(30) - ( 100cos(30) - Tension2*sin(22) - Ma A )μ = 0
OK, except the expression for the normal force is incorrect.
[EDIT: The right hand side should not be zero.]

My equation for tension1 was easier, just:
Tension2 = Mb A - Mb A
I think you have a couple of typographical errors here. The right-hand side reduces to zero.
Also, did you mean "Tension1" instead of "Tension2"?

I should just account for it as the same mass from before,
Yes
 
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  • #3
Woops, Thanks for the fast reply.
TSny said:
This equation says that NF = MaA which implies that the net force accelerating Ma is the normal force. Is that correct?
Yeah, that's what I was implying. Looking at the diagram, I was trying to account for the Y side.. and figured that the Y would be changing as well as it is not at rest..I edited the original post, I stupidly forgot the G.

Tension2*cos(22) - Magsin(30) - ( 100gcos(30) - Tension2*sin(22) - Ma A )μ = 0

Thanks for the reply!
TSny said:
I think you have a couple of typographical errors here. The right-hand side reduces to zero.
Also, did you mean "Tension1" instead of "Tension2"?

Yeah.. ment Tension 1.. stupid mistake.. wrote this all to fast
 
  • #4
Not sure what you mean by the "Y side" or that the "Y side would be changing". I don't see the Y direction indicated in the diagram.
 
  • #5
TSny said:
Not sure what you mean by the "Y side" or that the "Y side would be changing". I don't see the Y direction indicated in the diagram.

Sorry, Not sure what I ment by that. I put 100gcos(30) - Tension2*sin(22) - Ma A
in because I figured that when I calculated the normal force I would need to account of Acceleration as well since it is now moving.. now that I say that it doesn't make a ton of sense to me.. should I have just left it as 100gcos(30) - Tension2*sin(22)
 
  • #6
When the block accelerates along the incline, what is the component of acceleration in the normal direction?
 
  • #7
TSny said:
When the block accelerates along the incline, what is the component of acceleration in the normal direction?
Hmm, not quite sure what that means, would it be:

(100gcos(30) - Tension2*sin(22))/Ma= A
 
  • #8
Suppose we let the Y-axis be oriented perpendicular to the incline (that is, the Y direction is in the Normal direction).

What do you get if you set up Newton's 2nd law in terms of components in the Y direction: ΣFY = maY
 
  • #9
TSny said:
Suppose we let the Y-axis be oriented perpendicular to the incline (that is, the Y direction is in the Normal direction).

What do you get if you set up Newton's 2nd law in terms of components in the Y direction: ΣFY = maY
Oh, okay.. i see what your saying.. wouldn't it be something like:
100gcos(30) - (Tension2*sin(22) + 100gcos (30)

The tension and normal force acting in one direction, with the force of direction in the other? other wise I am a bit lost here haha

Thanks
 
  • #10
joemost12 said:
100gcos(30) - (Tension2*sin(22) + 100gcos (30)
I don't see what this represents. You have identified the forces acting on Ma. For each force, find an expression for the Y-component of the force. Also, think about the value of the Y-component of the acceleration of MA. Then you can set up the equation ΣFY = MaaY to find the normal force.
 
  • #11
TSny said:
I don't see what this represents. You have identified the forces acting on Ma. For each force, find an expression for the Y-component of the force. Also, think about the value of the Y-component of the acceleration of MA. Then you can set up the equation ΣFY = MaaY to find the normal force.

This was my Y component: 100gcos(30) - (Tension2*sin(22
What should the Y component be then? a little lost.. the axis are tilted.. should it be 100gsin(30) - Tension2*cos(22)
 
  • #12
joemost12 said:
This was my Y component: 100gcos(30) - (Tension2*sin(22
What should the Y component be then? a little lost.. the axis are tilted.. should it be 100gsin(30) - Tension2*cos(22)

Otherwise.. going to need some help here.. really not sure what the Y component could be
 
  • #13
Let's back up just a little to make sure we are on the same page. Can you list all of the forces that act on Ma?
 
  • #14
Could I instead use my original equation I used to solve for mass b instead? very lost..
 
  • #15
You know the general approach to problems that require the application of Newton's 2nd law:

1. Draw a free body diagram for each mass. Each diagram shows the forces on the mass as well as a choice of an x-y coordinate system.

2. Use each diagram to help set up the component form of Newton's second law: ΣFx = max and ΣFy = may for each mass.
(For a non-zero-mass pulley, set up Σ##\tau## = I ##\alpha##)

3. Solve the equations from step 2 for the unknowns.

So, we are trying to do step 1 for the block on the incline. In order to draw a free body diagram for this mass, you must first identify all of the forces acting on this mass. Your work for part (a) looks good. In part (b) you have essentially the same forces (but the values of some of the forces will be different) .

Can you show (or describe) the free body diagram for the block on the incline?
 
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  • #16
R8OEB.png


Okay, so here is what I think the force diagram is for the block, ignore the other stuff though.

So my equation for part 1 is Tension*cos(22) - Ma g sin(30) - ( Ma g cos(30) - Tension*sin(22) )μ = 0
sometimes I think about it such as the equation is parallel and perpendicular.

Would this be the Y equation:
Normal Force + Tension sin(22) - mgcos(30) = MaSo trying to think about how my equation would be different from the work in Part 1. Besides the fact that there is acceleration because it is not at rest.
Thanks for all your help on this, just a little lost on what from part 1 I need to do differently for part 2, other than account for acceleration
 
  • #17
Your diagram looks good for MA. As you say, you should think about the forces parallel and perpendicular to the incline. So, it is helpful to rotate the coordinate system so that the x-axis is parallel to the incline and the y-axis is perpendicular to the incline. Then when you set up ΣFx = max and ΣFy = may, you are working with the components parallel and perpendicular to the incline.
upload_2016-11-5_21-38-3.png


Try setting up ΣFy = may. This will allow you to get an expression for the normal force. Think about the value of ay.
 
  • #18
TSny said:
Your diagram looks good for MA. As you say, you should think about the forces parallel and perpendicular to the incline. So, it is helpful to rotate the coordinate system so that the x-axis is parallel to the incline and the y-axis is perpendicular to the incline. Then when you set up ΣFx = max and ΣFy = may, you are working with the components parallel and perpendicular to the incline.
View attachment 108543

Try setting up ΣFy = may. This will allow you to get an expression for the normal force. Think about the value of ay.

This would be my equation for may,
may * acceleration = Normal Force + Tension sin(22) - mgcos(30)

I guess I could only solve for the normal force within the whole system of equations, or I could put this in terms of the normal force and solve for this as Normal force in the other equation? If that is is the case I could do
Normal Force = may * acceleration - Tension sin(22) + mgcos(30)

And then plug into the parrallel equation:
Tension2*cos(22) - Magsin(30) - ( may * acceleration - Tension sin(22) + mgcos(30 )μ = Ma

right idea? or am I lost?
 
  • #19
Good. Definitely the right idea. In what direction does the block accelerate? So, what is the value of ay?
 
  • #20
TSny said:
Good. Definitely the right idea. In what direction does the block accelerate? So, what is the value of ay?
The block is going to accelerate up the ramp...
I guess to calculate Ay at this point I would use something like this:
Tnet = [ Tension2 - Tension1]*radius = .5 m r^2 ∝ ?
 
  • #21
In post #18 you wrote correctly that

may * acceleration = Normal Force + Tension sin(22) - mgcos(30)

Normal Force = may * acceleration - Tension sin(22) + mgcos(30)

We can drop the word "acceleration" since ay means the y-component of acceleration. So we can write

may = Normal Force + Tension sin(22) - mgcos(30)

Normal Force = may - Tension sin(22) + mgcos(30)

You can simplify this by noting the value of ay. Remember, that we chose the y-axis to be perpendicular to the incline. If you drew a vector representing the acceleration of the block, what would be the direction of that vector? What would be the y-component of that vector?
 
  • #22
TSny said:
In post #18 you wrote correctly that

may * acceleration = Normal Force + Tension sin(22) - mgcos(30)

Normal Force = may * acceleration - Tension sin(22) + mgcos(30)

We can drop the word "acceleration" since ay means the y-component of acceleration. So we can write

may = Normal Force + Tension sin(22) - mgcos(30)

Normal Force = may - Tension sin(22) + mgcos(30)

You can simplify this by noting the value of ay. Remember, that we chose the y-axis to be perpendicular to the incline. If you drew a vector representing the acceleration of the block, what would be the direction of that vector? What would be the y-component of that vector?
The y component would be sin() of it... the only thing i am confused about now

Now I have ay and ax though? Look I really appreciate your help but going through this step by step has become pretty tedious, what do you think the equation should look like?
 
  • #23
I suppose I could just say Asin(22) = Ay... I'm not sure at this point.. this problem has been so drawn out to the point wher ethis isn't helping much anymore..
 
  • #24
You are actually doing well with the problem. For some reason, though, you are having trouble seeing the numerical value of ay for the block on the incline, where the y-axis is perpendicular to the incline. This should be easy.

Suppose a crate accelerates along a horizontal floor. What would be the value of the vertical component of the acceleration of the crate?
 
  • #25
TSny said:
You are actually doing well with the problem. For some reason, though, you are having trouble seeing the numerical value of ay for the block on the incline, where the y-axis is perpendicular to the incline. This should be easy.

Suppose a crate accelerates along a horizontal floor. What would be the value of the vertical component of the acceleration of the crate?
oh.. I mean it would be 0, if the crate is remaining on the horizontal floor
 
  • #26
Exactly.

Apply this idea to the block on the incline.
 
  • #27
TSny said:
Exactly.

Apply this idea to the block on the incline.
Okay gotcha.. so I guess I can just account for the Ay as 0, meaning my normal force equation would be
Normal Force = - Tension sin(22) + mgcos(30)
as the 0 would cancel out the mass
 
  • #28
Yes. Good.
 
  • #29
I'm guessign at that point I could then use:
Tnet = [ Tension2 - Tension1]*radius = .5 m r^2 ∝ ?Tension2*cos(22) - Magsin(30) - ( - Tension sin(22) + mgcos(30) )μ = 0
Put this equation in terms of Tension2


solve the hanging block equation for t1
Mb a = T1 - Mb g
Tension1= Mb a + Mb g

then sub into this formula
Tnet = [ Tension2 - Tension1]*radius = .5 m r^2 ∝ ?
 
  • #30
joemost12 said:
I'm guessign at that point I could then use:
Tnet = [ Tension2 - Tension1]*radius = .5 m r^2 ∝ ?
Yes, this gives you another equation to work with. You should be careful with the signs. When the block accelerates up the incline, the pulley will rotate clockwise. So, it is a good idea to take clockwise rotation as the positive direction for torque and angular acceleration.


Tension2*cos(22) - Magsin(30) - ( - Tension sin(22) + mgcos(30) )μ = 0
Think about whether or not the right hand side should be zero.
 
  • #31
TSny said:
Yes, this gives you another equation to work with. You should be careful with the signs. When the block accelerates up the incline, the pulley will rotate clockwise. So, it is a good idea to take clockwise rotation as the positive direction for torque and angular acceleration.



Think about whether or not the right hand side should be zero.
Nope.. it shouldn't be.. it should be equal to Ma A, not 0
 
  • #32
Yes. It will be important to distinguish between the acceleration of block A and the acceleration of block B. So, I suggest writing the right side as MAaA. You can use the toolbar to enter superscripts and subscripts.
 
  • #33
Oh, I assumed the two blocks would have the same acceleration as they are all in the same system...

Shouldn't acceleration of A be equal to acceleration of B?
 
  • #34
joemost12 said:
Oh, I assumed the two blocks would have the same acceleration as they are all in the same system...

Shouldn't acceleration of A be equal to acceleration of B?

The two accelerations would be equal if the string on the left side of the pulley (connected to block A) were parallel to the incline. You will need to determine the relationship between aA and aB.
 

Related to How Does a Speck of Dust Affect the Motion of a Box on an Inclined Ramp?

1. What is a pulley?

A pulley is a simple machine that consists of a wheel with a groove around its circumference. It is used to change the direction of a force, typically to lift or move an object.

2. How does a pulley work?

A pulley works by utilizing the principle of conservation of energy. When a force is applied to one end of a rope or cable that is threaded through the groove of a pulley, the force is transmitted to the other end of the rope, causing the object attached to it to move.

3. What is an inclined ramp?

An inclined ramp, also known as an inclined plane, is a flat surface that is angled at a certain degree. It is used to reduce the amount of force needed to move an object to a higher or lower level.

4. How does an inclined ramp make work easier?

An inclined ramp makes work easier by reducing the amount of force needed to move an object. By increasing the distance over which the force is applied, the ramp decreases the amount of force required to lift or move an object.

5. How are pulleys and inclined ramps related?

Pulleys and inclined ramps are related in that they both use the principle of mechanical advantage to make work easier. Pulleys change the direction of a force, while inclined ramps reduce the amount of force needed to move an object. Both of these simple machines can be used together to create even greater mechanical advantage.

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