How Do You Calculate the Acceleration and Tension in a Pulley System?

[Mwyprism]

Homework Statement

A 98.6 kg box is resting on a horizontal surface. The coefficients of friction are uk= .1602 and us= .6027 for the surface to box. The box is attached to a 59.1 kg rock by a cord that passes over a frictionless pulley.

What is the acceleration of the 59.1 kg mass? Make sure the direction is correct.

What is the tension in the cord?

Homework Equations

Some equations I think may help with this problem (keep in mind I am clueless in Physics atm) are:

F=ma
(force of kinetic friction) Fkf = uk*Fn
(force of static friction) Fsf= us*Fn
Ft= mg+ma

The Attempt at a Solution

I have not enough time to type the 1/2 pg. of BS I started with.
My grade is slipping in this class and I need your help. Teacher doesn't give enough of it when I ask. I have one or more problems that I need help with similar to this, but will refrain from typing them incase I don't end up with any responses, as this is my first time using this forum.

 

[berkeman]

What does your free body diagram (FBD) look like? What are the forces on the box and on the rock due to the cord? What directions are those forces pointed in? If the forces are different, what does that tell you about the motion of the box and rock? Does it look like you should use uk or us? Why?

 

[Mwyprism]

My FBD for the box shows a small friction in the left direction and a longer arrow going to the right as the tension. For the rock's FBD, there is the force of gravity downwards and the force of tension upwards.
I would assume we would have to use both uk and us. But for the acceleration I'd guess we use Uk to find it, because the box would be moving (kinetic).
would the tension in the cord be:
Ft = mg + ma, we obviously would have needed to first find A which I am unsure of how to do, I'd further be unsure of what masses to use in the Ft equation. both mass of the rock and box or just one of them?

 

[berkeman]

Good start! You're close on the tension, but be careful to use separate mass symbols for the box and rock. And the tension on the rope from the box has two components -- some force from the acceleration of the box and some tension from the resistance of the friction. First determine if the whole system is moving or not. Assume no movement and see if the static frictional force will be overpowered or not...

 

[Mwyprism]

So you include both objects mass in the tension equation?

I think it would be best to answer the questions in order. So, we are looking for the acceleration of the 59.1kg rock. I tried something and got 9.8 m/s2 which I think is wrong since that is just the force of gravity. What I did was this:

A=?
m = 59.1kg
Fn (normal force) = mg
Fn = 59.1kg*9.8 m/s2
Fn= 579.18 N

F=ma
579.18N = 59.1kg*a
divided both sides by 59.1kg to get a by itself, and got 9.8 m/s2.

What did I do wrong?
Also, if you have AIM let me know.

 

[berkeman]

Well, the first thing you should do is multiply Fn by the statuc mu (us). What do you get? Why is understanding this force important to do first?

 

[Mwyprism]

That would give me 349.071 N but what does that represent? the minimum force it takes to get the box moving? how do I find the rock's acceleration?

 

[berkeman]

That would give me 349.071 N but what does that represent? the minimum force it takes to get the box moving? how do I find the rock's acceleration?

And what is the initial force supplied by the hanging rock via the rope and pulley? What does that tell you about what the box does in the beginning?

 

[Mwyprism]

The inital force supplied by the hanging rock is the rock's mass times gravity (579.18 N)? I don't know what that tells you about the box in the beginning. It is at rest? I feel like you're not answering my questions

 

[berkeman]

The inital force supplied by the hanging rock is the rock's mass times gravity (579.18 N)? I don't know what that tells you about the box in the beginning. It is at rest? I feel like you're not answering my questions

I'm not directly answering questions that you should be able to answer yourself, given proper hints. Now go back and re-calculate the force due to static friction at the box. I don't get 349N for that. Compare the static friction force available to the weight of the rock.

 

[Mwyprism]

If you didn't get 349 N, tell me how you came to your answer, because obviously I don't know how to find the force due to static friction at the box. I tried, now help because I don't know what to do. I got 349 N by using:

Ffs = Us*Fn
Ffs = Us*mg
Ffs = .3027*59.1kg*9.8m/s2
= 349

 

[berkeman]

The box is 98.6kg, and the rock is 59.1kg.

 

[Mwyprism]

Are you saying I just used the wrong mass?
If so, it would be 582 N.

 

[berkeman]

And re-checking your post #9 tells you what? Look back at your FBD if that helps. Looks like you're done!

Well, except for the obvious two quiz questions I'll ask when you post the answer...

 

[Mwyprism]

I'm really confused, how can I be done? I don't know what 582 N is. It's the force needed to move the box? How does that answer the question What is the acceleration of the 59.1kg mass/rock?

 

[berkeman]

Okay, to summarize some of the steps you've taken, fill in the following:

-- Mass of box, force of static friction available before box starts slipping.

-- Mass of rock, force of tension available by rock pulling down on rope, force of tension available to start the box slipping.

-- Conclusion about acceleration of the box and rock.

Ready for the two quiz questions now?

 

[Mwyprism]

I wish you'd stick around longer and help me until I'm actually finished, this has been really prolonged and I have a ton of homework, I do appreciate your help so far though

 

[berkeman]

Okay, to summarize some of the steps you've taken, fill in the following:

-- Mass of box, force of static friction available before box starts slipping.

-- Mass of rock, force of tension available by rock pulling down on rope, force of tension available to start the box slipping.

-- Conclusion about acceleration of the box and rock.

Ready for the two quiz questions now?

Please fill in the blanks, and I think you're done.

 

[Mwyprism]

98.6 kg is the mass of the box.
582 N is the force of static friction avail. before box starts slipping.
59.1 kg is the mass of rock.
I don't know how to find the force of tension available ??
I haven't even found the acceleration of the rock !

 

[berkeman]

The initial tension before anything starts to move is the weight of the rock. What is that. And then what happens?

 

[Mwyprism]

The initial tension is 59.1kg. and then what? I don't know, you tell me haha. I don't have any equation to find tension dealing with the rock's mass. I believe I need to find the acceleration of the rock before I find tension

 

[Mwyprism]

Please stop going away, I need to sign off soon

 

[berkeman]

I'm at work, sneaking moments here and there to check the updates in the forums. Honestly, I've done about all I can do to help you figure this out, short of telling you the answer outright. I have to go soon as well. Please just finish answering the last few leading questions that I've asked. If the system starts at rest, what happens next?

 

[Mwyprism]

You're ignoring my questions when I'm clearly completely lost. You're seriously going to leave after all of this trying I have done, without telling me how to get the answer??

 

[cristo]

If berkeman goes before you've finished.. I'm sure someone else will step in!

 

[Mwyprism]

well it has been all this time and nobody else has cared to step in. and I don't have all night, plus I'm exhausted and frustrated.

 

[cristo]

It's a bit rude to butt in when he was obviously half way through helping you (he says, butting in!)

 

[berkeman]

You're ignoring my questions when I'm clearly completely lost. You're seriously going to leave after all of this trying I have done, without telling me how to get the answer??

Yes. I've told you the answer at least four different ways, and you aren't answering my leading questions. One last try, and I'm heading home.

Look at what you have so far:

-- The box can supply up to 582.38N of horizontal resistive force before it will start to slip and move.

-- The rock weighs 579.18N.

-- What is the tension in the rope if nothing is moving at first? The weight of the rock is pulling down on the rope at one end --> that tells you the tension in the rope transferred to the horizontal pulling force on the box (via the pulley).

-- Is the force pulling down on the rope (and thus pulling sideways on the box) enough to get the box slipping?


PS -- Just saw christo's note. Thanks dude! I'm out.