physics problem 3...PLEASE HELP ME

[piper210_355]

Like i sayed in my other posts this is probably easy but my teacher doesn't teach us anything so i don't have a clue how to do this. Can someone please explain how to do it?

 

A window washer pushes a sponge up a vertical window at a constant speed by applying a force. The sponge has a mass of .800 kg, and the coefficient of kinetic friction between the sponge and window is u_k=0.253. Determine the magnitude of the applied force, and the magnitude of the normal force exerted by the window on the sponge.

 

*There is a 20 degree angle

Edited October 9, 2008 by piper210_355
multiple post merged

[Klaynos]

First of all how many forces are involved and can you write down their values (in terms of other quantities)?

[big314mp]

Something doesn't seem entirely right with this question...

[Klaynos]

Something doesn't seem entirely right with this question...

 

There doesn't look to be enough info does there... I thought it was just me missing something because I hadn't worked threw it...

[piper210_355]

That's all the info that is given in the problem

there is a picture, but it only shows that there is a 20 degree angle, if that's what your talking about.

I added that to the post though

[big314mp]

That is exactly what was needed.

 

Remember that the normal force is perpendicular to the surface (in this case the window).

 

There is also a force moving the sponge upwards.

 

These two forces come from the force applied by the window washer (think vector components), and they exist in a constant ratio with each other (do the trig to find the ratio).

 

Remember that the sponge is moving up.

[piper210_355]

ok i think that i found an equation that could help me with this:

F(f)=uF(normal)

 

But how would i find the normal force?

in another problem it was the mass times gravity, but in that problem the object was on the ground. Would it be different since it is vertical?

 

or does that only apply to the vector components

F=F(x)+F(y)+F(z)

such that F(y) isn't 0 now, but instead it is the kinetic friction maybe?

 

i don't know. I think i might be getting closer to getting it but i'm not completely sure.

[big314mp]

Remember that the normal force is the force perpendicular to the surface.

 

When an object is sitting on the ground, this force is the weight (since that force goes perpendicular to the surface) of the object.

 

Now what if I push an object against a wall? The force that I apply will be perpendicular to the wall, and the force of my push will be the normal force, not the object's weight.

 

In your case, you have a window washer, pushing an object against a window. But he isn't pushing the object perpendicular to the window. However, one of the components of the force he applies is perpendicular to the window. The other, of course, is parallel to the window.

 

You will have two variables, so you will need two equations, btw.

[piper210_355]

I'm guessing that one of the equations is the F(friction)=uF(normal).

In another problem that I did, the normal force was mass x gravity (although I'm not sure if i can do that for this prob since it is vertical).

If i can do this then it would be

 

F(f)=(.253)[(.800)(9.81)]

F(f)=1.98

 

Would this be the magnitude of the normal force or am I way off?

 

I'll do some more research to try to find an equation that will help me find the magnitude of the Force applied.

Edited October 12, 2008 by piper210_355

[big314mp]

We are getting closer

 

F(f) = u*F(n) is an equation that you will have to use.

 

Remember that the sponge is moving upwards at constant speed. This means that the window washer is lifting the sponge, and so must apply enough force to lift the weight of the sponge. He must also overcome the frictional force.

 

Since this problem is on a vertical surface, the normal force is not going to be the weight of the object. The normal force will come from the window washer.

 

The window washer is pushing at a 20 deg angle. This means that his force is being applied to the sponge at a 20 deg angle. To find the forces you need, you must break this force into components. One component will be perpendicular to the window. This is your normal force (see the examples I posted earlier). The force going parallel to the window must overcome (think about what you can set this force equal to) the frictional force and the weight of the sponge.

[piper210_355]

so if i break it up into components would it be

 

F= F(normal)+F(frictional)+weight of sponge

 

???

I think i'm starting to get it a little bit more but I'm still a bit confused

[big314mp]

Closer...

 

F(vertical) = F(frictional) + weight of sponge

 

Now you need to figure out how to get F(f).

 

You also need a way to relate F(y) to F(x).

 

You're doing great!

[piper210_355]

wouldn't i find F(f) by using the equation from earlier?

In order to do this i need to find the normal force.

Can i somehow use the angle (maybe by using sin or cos) to get the normal force

 

and i was thinking wouldn't the applied force and the normal force equal the same thing?

If so then i only have to look for one and i'll have both.

[big314mp]

You do use trig to get the normal force, but you will be best off using the tangent function.

 

I'm going to define the force perpendicular to the window as F(x) and the force parallel to the window as F(y). I'll define the applied force as F(a). Thus, F(x) and F(y) are the vector components of F(a).

 

F(x) will be the normal force used to calculate friction. F(y) is the force that moves the sponge up the window (against gravity, and against friction). F(a) does not equal F(x), but you need to know either F(x) or F(y) to calculate F(a), using the sine or cosine function.

[piper210_355]

ok so i calculated the sin(20)=.342

and the cos(20)=.940

[big314mp]

F(y) = weight of sponge + Frictional force

 

F(f) = u*F(x)

 

Now you need a trig function that will relate F(y) to F(x) using the angle given in the problem.

[piper210_355]

actually on my paper it looks like this

would it be the same thing?

Edited October 12, 2008 by piper210_355

[big314mp]

No, it isn't the same. Where I have 20 deg, it should be 70 deg. Everything else should be the same.

[piper210_355]

ok so what i did was that i took the sin 70 which if i'm not mistaken should equal F(x).

sin 70=F(x)

.94=F(x)

 

Then I put this in the equation for frictional force that you gave me earlier.

F(f)=uF(x)

F(f)=(.253)(.94)

F(f)=.238

 

After that I put it in to the equation you gave for F(y).

F(y)=weight of sponge + F(f)

F(y)=(.8)+(.238)

F(y)=1.048

 

My plan is to do the pythagorian therum to find the applied force, but I want to make sure i am doing it right so far.

[big314mp]

sin(70) = F(x)/F(a), or F(a)*sin(70) = F(x)

 

Remember that sine is opposite over hypotenuse.

 

Replace your .94 with F(a)*sin(70), and you will get:

 

F(y) = F(a)*sin(70)*.253 + .8*9.8

 

Now you have an equation with two variables, and you need it to be in one variable. BTW, the 9.8 comes from converting kg to newtons, as the kg is not a unit of weight, it is a unit of mass.

 

So you need to find a relation between F(y) and F(a):

 

cos(70) = F(y)/F(a), or F(a)*cos(70) = F(y)

 

Substitute this in for F(y), and you will have an equation in one variable, which can then be solved using algebra.

 

While you can do this problem using sine and cosine, on a test you may want to try the tangent function, as it can often get you to your answer in a more straightforward route. If you decide to do that, make sure you practice it.

[piper210_355]

ok so i substituted it in for F(y):

F(a)cos70=F(a)(sin70)(.253)+(.8)(9.8)

.34F(a)=.238F(a)+7.84

.102F(a)=7.84

F(a)=76.86

 

 

Then I used this to find the value of F(x) since that's the normal force.

 

F(x)=F(a)sin70

F(x)=(76.86)(sin 70)

F(x)= 72.22

 

Thank you sooooo much for your help

I'm sorry it took so long

I would have never been able to do these probs w/o your help