What Force Does the Table Exert on a Box Attached to a Pulley System?

[kaytee28]

A box weighing 71 N rests on a table. A rope tied to the box runs vertically upward over a pulley and a weight is hung from the other end



Determine the force that the table exerts on the box if the weight hanging on the other side of the pulley weighs each of the following.
(a) 27 N

(b) 66 N

(c) 98 N


help meeee?

 

[cristo]

This isn't relativity! You should post in the homework forums in future. Also, note that we cannot help you until you have shown us some effort. So, what have you tried thus far?

 

[kaytee28]

Okay...I'll post it in the homework one next time. I'm sorry about that.. And I've tried multiplying all of them by gravity and it's gotten me no where and I don't know what else to do for it.

 

[malty]

An attempt would be nice to see, anyways is the pulley perfectly frictionless or can we assume it to be the case.

This problem just requires the sum of the forces acting upon the table. Calculate those in each case and you have your answer.

Hint:Draw a diagram, and look at what the mass added to the pulley would do.

 

[cristo]

Take each part separately, and try drawing a free body diagram for the box on the table. What forces are acting on the box?

 

[kaytee28]

We just have to assume about the friction being there, but what do you mean by just the sum of all of the forces?

And normal force, tension, friction, acceleration, velocity, and acceleration are all acting on the box.

 

[malty]

We just have to assume about the friction being there, but what do you mean by just the sum of all of the forces?

And normal force, tension, friction, acceleration, velocity, and acceleration are all acting on the box.

Well, I'd say you'd had to assume that there's no friction in the pulley if no details are given. As cristo said draw the diagram, and see what is happening. . .

 

[kaytee28]

okay, then there is no friction, but I drew the diagram but I still am confused about how to figure the problem out and plugging things in where.

 

[malty]

okay, then there is no friction, but I drew the diagram but I still am confused about how to figure the problem out and plugging things in where.

Any chance you could show us this diagram?

ok consider the block with the pulley what forces are acting on it, and in what direction?

Secondly consider the mass hanging down, again what forces are acting on it, and in what direction.
Now what difference does the pulley actually make?

 

[kaytee28]

I don't know how to show the diagram on here, otherwise I would. On the box..wouldn't normal force and tension being going upwards, gravity going downwards? And the mass would have gravity and acceleration going down? I'm not really sure about the difference it makes.

 

[malty]

I don't know how to show the diagram on here, otherwise I would. On the box..wouldn't normal force and tension being going upwards, gravity going downwards? And the mass would have gravity and acceleration going down? I'm not really sure about the difference it makes.

The pulley is the key, what difference does that make.
Hint : Directions

 

[kaytee28]

I don't get what difference it makes at all. I'm so confused on how to do this one.

 

[malty]

I don't get what difference it makes at all. I'm so confused on how to do this one.

Ok, what would the pulley do, if no box was attached to it, and only the masses you were to add to it were attached to it?

What would happen to the masses?

Now consider what would happen if the masses were removed and only the box was attached to the pulley.. what would happen to the box?

 

[kaytee28]

If there was no box, the weight would move until it hit the ground.

And if there was only a box, it would be resting on the table and wouldn't be moving anywhere.

 

[malty]

If there was no box, the weight would move until it hit the ground.

And if there was only a box, it would be resting on the table and wouldn't be moving anywhere.

Now put the two of these together. . .

Hint: For the box to be in equilbrium and stationary Net F=0

 

[kaytee28]

but then what equation would I use for that if F=0? because then it wouldn't come out right?

 

[malty]

but then what equation would I use for that if F=0? because then it wouldn't come out right?

No,
The Net Force = 0

 

[kaytee28]

but then what equaton would I use if F=0? because then wouldn't it not work out correctly?

 

[kaytee28]

so then what equation does that go into for net force? because then I have to do what the table exerts on to each box.

 

[malty]

but then what equaton would I use if F=0? because then wouldn't it not work out correctly?

no the net force equals the sum of the forces acting on the box.

i.e the net force acting on the chair you're sitting on at the mo (i presume you're sitting down) are
(your weight)+ (the force exerted upon the chair by the ground) = 0
mg+f(ground)=0

 

[malty]

so then what equation does that go into for net force? because then I have to do what the table exerts on to each box.

Let's go back to your pulley, and consider all forces, and their directions acting on the box; the sum of these forces must equal zero.

 

[kaytee28]

but what would the 'ground' be because there is nothing pushing it or anything?

 

[malty]

but what would the 'ground' be because there is nothing pushing it or anything?

In your question its the table, and there are forces acting on it, that's what you've got to find, you have all the necessary info.

 

[kaytee28]

I'm just confusing myself on the equation and everything..

 

[malty]

I'm just confusing myself on the equation and everything..

Okay let's go back to the start.

Calculate the force the table is exerting on the box without any mass/ pulley present.
Now
You've deduced that without the pulley,
the masses would fall to the floor (quantify this)

For the box to remain in equilibrium the NET FORCE MUST EQUAL ZERO.

. . .

 

[kaytee28]

I don't understand how to get the force that the table is exerting on the box..

 

[malty]

I don't understand how to get the force that the table is exerting on the box..

Again I'll say that THE NET FORCE MUST EQUAL ZERO.

What exactly does that mean?
If there are only two forces acting on a object, and these forces opposite the direction, then magnitude (size) of these forces must be equal.
F1+(-F2)=o
=> F1=F2