Finding largest value of H of a ramp, from a FBD

[Alison A.]

Homework Statement

The dimension h is to be determined to that a worker can comfortably slides boxes weighing up to 106 lb up and down a frictionless incline. If the worker can apply a 50 lb horizontal force to the box, what is the largest value h should have?

Hint given: Before taking any other action, draw a FBD of the box. Isolating the box from its surroundings requires a closed surface passing between the worker's hands and the box as well as between the bottom of the box and the incline. It may be more natural to define and solve for an angle of inclination α and then determine h from tanα = h/6.

Picture and force body diagram attached
http://imgur.com/a/EBBCi

Homework Equations

ΣFx=0
ΣFy=0
tanα = h/6

The Attempt at a Solution

I am having a hard time differentiating between the force of 50lb along the ramp, and then the distance 6ft since they are not the same type of thing. Do I need to find the distance of the 50lb force? Or the force of 6ft? Because from there I could solve for α.

I am not sure if I am calculating a system that's in equilibrium since he is actively pushing the box up.

 

[stockzahn]

First of all, I'm not sure what comfortably means in this case. However your FBD isn't correct (why is there a right angle between your forces?), try to think about it again. Maybe it is easier, if you draw the box with the forces and without the ramp and the worker.

Regarding the equilibrium: Think about Newton's first law. What does it say about objects in movement?

 

[Alison A.]

First of all, I'm not sure what comfortably means in this case. However your FBD isn't correct (why is there a right angle between your forces?), try to think about it again. Maybe it is easier, if you draw the box with the forces and without the ramp and the worker.

Regarding the equilibrium: Think about Newton's first law. What does it say about objects in movement?

I went to my engineering learning center here on campus and one of the coaches said that my FBD was correct... I kind of casually added that right angle when I was trying to figure something out, so ignore that right angle sorry! But are other things wrong with it?

Since an object in motion tends to stay in motion in the same direction unless acted upon by an unbalanced forced I'm going to conclude that this system is not in equilibrium.

 

[stockzahn]

I went to my engineering learning center here on campus and one of the coaches said that my FBD was correct... I kind of casually added that right angle when I was trying to figure something out, so ignore that right angle sorry! But are other things wrong with it?

It's hard to say, if I don't see the correct FBD, but based on your drawing: In which direction does the weight force of the box point (or what's the angle between the force of the worker and the weight force of the box)? Then it's necessary to figure out what part (component) of the weight force points "against" the force of the worker (which should not exceed his strength). For this reason it's helpful to split up the weight force of the box in two components - parallel to the ramp and rectangular to it.

Since an object in motion tends to stay in motion in the same direction unless acted upon by an unbalanced forced I'm going to conclude that this system is not in equilibrium.

What if the force used by the worker to push the box (apart from the first acceleration, which must be higher) corrisponds to the part of the weight force of the box which is parallel to the ramp?

 

[Alison A.]

It's hard to say, if I don't see the correct FBD, but based on your drawing: In which direction does the weight force of the box point (or what's the angle between the force of the worker and the weight force of the box)? Then it's necessary to figure out what part (component) of the weight force points "against" the force of the worker (which should not exceed his strength). For this reason it's helpful to split up the weight force of the box in two components - parallel to the ramp and rectangular to it.

If I don't know the angle to anything, how do I find the components. Does the dimension of 6ft come into play anywhere here?

What if the force used by the worker to push the box (apart from the first acceleration, which must be higher) corrisponds to the part of the weight force of the box which is parallel to the ramp?

Then yes it would be in equilibrium.

 

[stockzahn]

If I don't know the angle to anything, how do I find the components. Does the dimension of 6ft come into play anywhere here?

Take α as your (variable) angle - it is crucial for the solution. The 6 ft are only necessary to calculate h. If it wouldn't be 6 ft, but 60 or 600 ft, solution for α will still be the same - only h is varying with the length of the ramp. The weight force on the other side is important in the first place.

Imagine what would happen if you decrease or increase α. What would happen if α = 90°? What with 0°? In which of these cases the worker would have to push harder to move the box? What happens with angles between these two values? Can you find a (qualitative) law, that describes what the force of the worker has to be depending on the angle α?

Then yes it would be in equilibrium.

Correct. There could be an equlibrium - when the force of the box parallel to the ramp is the sam as the force used by the worker.

 

[Alison A.]

Take α as your (variable) angle - it is crucial for the solution. The 6 ft are only necessary to calculate h. If it wouldn't be 6 ft, but 60 or 600 ft, solution for α will still be the same - only h is varying with the length of the ramp. The weight force on the other side is important in the first place.

Alright, that makes sense.

Imagine what would happen if you decrease or increase α. What would happen if α = 90°? What with 0°? In which of these cases the worker would have to push harder to move the box? What happens with angles between these two values? Can you find a (qualitative) law, that describes what the force of the worker has to be depending on the angle α?

It would be harder to push the box when the angle increases and easier when the angle decreases. I'm not really sure where to derive a law from that. As the angle increases the force needed to push the box up also increases? Vice versa?

 

[haruspex]

There is a simple equation relating alpha, h and the distance 6ft. What is it?
Put that aside for later and just work with alpha now. Since the box is not accelerating up the incline (except slightly to get started), it is effectively the same as a statics question. Write out the usual force balance equations using the three forces and angle in your FBD.

 

[Alison A.]

After having help on another problem I realize that the FBD should probably look something like this

Not sure where α would be located. I think it would be beside the normal vector in respect to the y-axis.

Finding the components of this I got
ΣF_x=50-Ncos(α)=0
ΣF_y=-106-Nsin(α)=0

However when I try to solve for α I get very strange numbers.

 

[haruspex]

After having help on another problem I realize that the FBD should probably look something like thisView attachment 88805
Not sure where α would be located. I think it would be beside the normal vector in respect to the y-axis.

Finding the components of this I got
ΣF_x=50-Ncos(α)=0
ΣF_y=-106-Nsin(α)=0

However when I try to solve for α I get very strange numbers.

Check the signs in your equations.
Also, what you wrote about where alpha should be in your diagram was correct, but your equations tell a different story.

 

[Alison A.]

ΣF_x=50-Ncos(α)=0
ΣF_y=-106+Nsin(α)=0

Solving for them I got N to be ≈ 117.2 making α ≈ 64.7°
I don't think that's right because I keep getting different values when I plug in N for α in both equations

 

[haruspex]

ΣF_x=50-Ncos(α)=0
ΣF_y=-106+Nsin(α)=0

Solving for them I got N to be ≈ 117.2 making α ≈ 64.7°
I don't think that's right because I keep getting different values when I plug in N for α in both equations

That is the correct solution to those equations, but did you see my edit to my last post? You wrote that alpha should be the angle between the normal and the y axis. That is correct, but it is not what your equations say.

 

[Alison A.]

That is the correct solution to those equations, but did you see my edit to my last post? You wrote that alpha should be the angle between the normal and the y axis. That is correct, but it is not what your equations say.

Hm..
The normal vector components aren't -Ncos(α)+Nsin(α)?

 

[haruspex]

Hm..
The normal vector components aren't -Ncos(α)+Nsin(α)?

They are, but you need to get them the right way round!
I find it helpful to check I have this right by considering the case where the angle is zero. When alpha is zero, will N act entirely in the X direction or the Y direction? Is that what your equations say?

 

[Alison A.]

They are, but you need to get them the right way round!
I find it helpful to check I have this right by considering the case where the angle is zero. When alpha is zero, will N act entirely in the X direction or the Y direction? Is that what your equations say?

When alpha is zero it would be act completely in the X direction? And my equation doesn't say that

 

[haruspex]

When alpha is zero it would be act completely in the X direction? And my equation doesn't say that

No, when alpha is zero the ramp would be horizontal. That makes the normal force vertical, i.e. the y direction.
When alpha is zero, cos is 1 and sin is zero.

 

[Alison A.]

No, when alpha is zero the ramp would be horizontal. That makes the normal force vertical, i.e. the y direction.
When alpha is zero, cos is 1 and sin is zero.

So only Nsin(α)?

 

[haruspex]

So only Nsin(α)?

What do you mean? What is 'only Nsin(α)'?
All I am asking you to check is whether you have sin and cos crossed over in your latest equations. I.e. should Nsin(α) be in the X equation or in the Y equation? When alpha is zero, that value will be zero.

 

[Alison A.]

ΣFx=50-Ncos(α)+Nsin(α)=0
ΣFy=-106=0?

I get that when the ramp would be flat it would have no y-component

Oh wait, so
ΣFx=50+Nsin(α)=0
ΣFy=-106-Ncos(α)=0

Solving for N and α= 117.2 and 25.25°?

Yup, I then solved for h and got it right. Phew.

Thank you so much for all your help again, it's 3:30am I think I'll go to bed now haha.
Goodnight!

 

[haruspex]

ΣFx=50-Ncos(α)+Nsin(α)=0
ΣFy=-106=0?

I get that when the ramp would be flat it would have no y-component

Oh wait, so
ΣFx=50+Nsin(α)=0
ΣFy=-106-Ncos(α)=0

Solving for N and α= 117.2 and 25.25°?

Yup, I then solved for h and got it right. Phew.

Thank you so much for all your help again, it's 3:30am I think I'll go to bed now haha.
Goodnight!

Bingo. Sweet dreams.

 

[Awdswagon]

How did you solve for N?

 

[haruspex]

How did you solve for N?

Nsin(α) and Ncos(α) are known. What equation relates sin(α) and cos(α)?

 

[Awdswagon]

Tan=sin/cos Right? Does N represent the normal force?

 

[Awdswagon]

Thank you for replying so promptly too by the way! I appreciate the help.

 

[Awdswagon]

I figured it out!