Writing Force and Tension Equations

[MysticDude]

Homework Statement

All answers shuold be in terms of m,g, θ ,μ_k and μ_s
1.Write the equation used to calculate the force required to move a block of mass m across a floor with a constant velocity. [SOLVED]
2.Write the equation to calculate the tension in a rope holding a block of mass m motionless on a ramp. [SOLVED]
3.Write the equation used to calculate the tension in a rope that is dragging a block of mass m up a ramp with constant acceleration.[SOLVED]
4Write the equation used to calculate net force on a block of mass m that is accelerating down a ramp.

Homework Equations

Well, you are supposed to write the equations :P

The Attempt at a Solution

1.Okay for number 1, since the velocity is constant the acceleration is 0. In other words F - f = 0 where f is the friction force. So the force will be F = f. and since everything has to be in the terms as the provided instructions tell us, my final answer will be F = μ_smg.

2. Here is where I stopped and got a little confused. I think that the force is just mgsin(θ). In other words I think that my final answer is F = mgsin(θ).

3. I am stuck here :P.Note: As we progress, I will put [SOLVED] around the solved questions.
Thanks for any help!

 

[Redbelly98]

1.Okay for number 1, since the velocity is constant the acceleration is 0. In other words F - f = 0 where f is the friction force. So the force will be F = f. and since everything has to be in the terms as the provided instructions tell us, my final answer will be F = μ_smg.

Since the problem states the block would move(←key word!) at a constant velocity, can you think of a simple modification to your answer?

2. Here is where I stopped and got a little confused. I think that the force is just mgsin(θ). In other words I think that my final answer is F = mgsin(θ).

That's correct if there is no friction. Does the problem indicate whether friction is present?

3. I am stuck here :P.

Is it supposed to be constant acceleration or constant velocity? Either way, you should start by drawing a free body (force) diagram.

 

[MysticDude]

Since the problem states the block would move(←key word!) at a constant velocity, can you think of a simple modification to your answer?That's correct if there is no friction. Does the problem indicate whether friction is present?Is it supposed to be constant acceleration or constant velocity? Either way, you should start by drawing a free body (force) diagram.

So for number 1, my answer should be F = μ_kmg right? Kinetic friction.

For number 2, I don't know whether or not there is friction because I only have as much information as is given in my problem.

For number 3, I drew a simple free body diagram of the object on the ramp using Gimp. [PLAIN]http://img153.imageshack.us/img153/4083/nov22hwnum3.jpg

 

[Condensate]

Do you mind if I pitch in? :)

So for number 1, my answer should be F = μ_kmg right? Kinetic friction.

Looks right, to me at least!

For number 3, I drew a simple free body diagram of the object on the ramp using Gimp. [PLAIN]http://img153.imageshack.us/img153/4083/nov22hwnum3.jpg[/QUOTE]
Is the force of the ramp on the box going to be directly opposite mg?

 

[MysticDude]

Do you mind if I pitch in? :)

Looks right, to me at least!Is the force of the ramp on the box going to be directly opposite mg?

I'm not sure what you mean by the second part of your reply. Since mg is the gravitational force, it should always point directly downwards right? Or are you talking about the normal force and how it should be perpendicular to the ramp? I still don't have an idea on the equation though. :(

Oh and thanks for helping me!

 

[Condensate]

I'm not sure what you mean by the second part of your reply. Since mg is the gravitational force, it should always point directly downwards right? Or are you talking about the normal force and how it should be perpendicular to the ramp? I still don't have an idea on the equation though. :(

Oh and thanks for helping me!

Yep, the normal force should be perpendicular to the ramp, and its magnitude will depend on that of the Y component of mg (this is after re-orienting the picture so that you end up with three axis-aligned forces and mg as your only slanted force.)

And no problem (about the help)!

 

[MysticDude]

Yep, the normal force should be perpendicular to the ramp, and its magnitude will depend on that of the Y component of mg (this is after re-orienting the picture so that you end up with three axis-aligned forces and mg as your only slanted force.)

And no problem (about the help)!

Ok so it should be something like this If I don't rearrange the axes right: [PLAIN]http://img580.imageshack.us/img580/981/nov22hwnum32.jpg

 

[Condensate]

Ok so it should be something like this If I don't rearrange the axes right: [PLAIN]http://img580.imageshack.us/img580/981/nov22hwnum32.jpg[/QUOTE]
Yeah, so everything after that in terms of finding the equation of #3 should be smooth sailing.

 

[MysticDude]

Ok so after rearranging the axes I get this:
[PLAIN]http://img143.imageshack.us/img143/8599/nov22hwnum33.jpg

Normally if mg was directly downward, the equation would have been F -μ_kF_{n(where is the normal force or mg) = ma. But now mg is slanted by a θ west from south, I think that the formula now should be F - μkmgsin(θ) = ma. What do you think about this?}

 

[Condensate]

Ok so after rearranging the axes I get this:
[PLAIN]http://img143.imageshack.us/img143/8599/nov22hwnum33.jpg

Normally if mg was directly downward, the equation would have been F -μ_kF_{n(where is the normal force or mg) = ma. But now mg is slanted by a θ west from south, I think that the formula now should be F - μkmgsin(θ) = ma. What do you think about this?}

_{If θ is west from south, then yeah, I think I agree with that answer too. :)
So F= ma + μkmgsin(θ) ?}

 

[MysticDude]

If θ is west from south, then yeah, I think I agree with that answer too. :)
So F= ma + μkmgsin(θ) ?

WooHoo I finally got it. I was thinking that it was going to be really complex and what not, but you showed me the light :P. By doing some adding I get what you get :P.

 

[Condensate]

WooHoo I finally got it. I was thinking that it was going to be really complex and what not, but you showed me the light :P. By doing some adding I get what you get :P.

Yeah moving terms from one side of the equation to the other is a pain, and if you ever need help with it again, I think I'll take one for the team. :P
10 (:P) s

You're good with question 4 now, too?

 

[MysticDude]

Yeah moving terms from one side of the equation to the other is a pain, and if you ever need help with it again, I think I'll take one for the team. :P

You're good with question 4 now, too?

I haven't started number 4 yet. But I think that this one is easier. But that's me :P I can be wrong hehe.

 

[Condensate]

Wait, MysticDude. The X component of mg was overlooked in problem 3.
My bad.

 

[MysticDude]

Wait, MysticDude. The X component of mg was overlooked in problem 3.
My bad.

Once you said that I visually understood it. If we use θ west from south, then it has to be cos(θ) right? So technically we are right if you use θ is from 0° to the line of mg right?

 

[Condensate]

Huh? Sorry, maybe I'm just misunderstanding your post.

But yeah, the equation described earlier in the thread was missing that cos term.

 

[MysticDude]

I'll get to number 3 later because I think we are correct so, can we work on number 4 please? I'm not rushing since I have the whole week to finish this homework but I also do want to get it over with.