Can someone explain the inclined plane picture to me?

[Lothar]

Hello everyone,
Right now I'm working on an inclined plane problem.

"a) The worker exerts a force(F) of 86 N. How much work does he do?

(b) How much work is done by gravity? (Be careful with the signs you use.)

(c) The coefficient of friction is µ = 0.20. How much work is done by friction? (Be careful with the signs you use.)
"

It gives me information about side lengths and masses. Really, I think I should be able to do this on my own, I just don't understand the diagram very well. The picture we've used in class is very similar to this one:
http://dunningrb.files.wordpress.com/2007/10/300px-free_bodysvg.png

Could anyone explain to me what the lines mean?
Thanks

 

[dx]

What lines? You mean the arrows? They are the force vectors. The one marked mg is the weight of the body, the ones with sin and cos are the components of the weight along the plane and perpendicular to it, the one marked f is the friction and the one marked N is the normal force exerted by the inclined plane on the body.

 

[Hootenanny]

It gives me information about side lengths and masses. Really, I think I should be able to do this on my own, I just don't understand the diagram very well. The picture we've used in class is very similar to this one:
http://dunningrb.files.wordpress.com/2007/10/300px-free_bodysvg.png

Could anyone explain to me what the lines mean?
Thanks

The arrows represent forces or components of forces. The arrow labelled N denotes the normal reaction force, the arrow labelled mg denotes the weight of the box and the arrow labelled f represents the applied force or friction if friction is present.

mgsinθ represents the component of the weight acting parallel to the slope whereas mgcosθ represents the component of the weight that acts perpendicular to the slope.

Edit: dx beat me to it

 

[Lothar]

work = FD so which one of the componets would I use to find work done by gravity?
mgsinθ or mgcosθ. I thought it was sin, but I'm getting an incorrect answer.
Triangle has a hyp of 5 and a opposite side of 3m. The angle is 36.87 degrees. The crate weighs 95N
95sin(36.87) = 57 J. What am I doing wrong?

 

[dx]

You forgot to multiply by D.

 

[Lothar]

I noticed that after I posted. I've tried with 171 (opposite side) and 285 (hyp) and neither are the correct answer.

 

[dx]

Didn't you say the hypotenuse was 5? Also, you have to convert the length into meters first to get the answer in Joules. In what units are the lengths given?

 

[Lothar]

Yeah the hyp was 5 meters.
5m*95N*sin(36.87) = 285

 

[dx]

How far up the plane does the worker move the box? Is it given in the question?

 

[Lothar]

Question:
A worker pushes a crate weighing W = 95 N up an inclined plane. The worker pushes the crate horizontally, parallel to the ground, as illustrated in Figure 10-21
Picture:
http://i19.photobucket.com/albums/b188/Luigio97/physics.gif

(a) The worker exerts a force(F) of 86 N. How much work does he do?

(b) How much work is done by gravity? (Be careful with the signs you use.)

(c) The coefficient of friction is µ = 0.20. How much work is done by friction? (Be careful with the signs you use.)

 

[dx]

Your answer was correct but you had the wrong sign. The parallel component of the weight is pointing opposite to the direction in which the box is being moved, so it is -285 J.

 

[Lothar]

Oh. That makes sense. The positive direction would be up the ramp, so since the gravity would bring it down then the sign would be negative.

Do I need to use one of the components for part a? The given force times the given hyp or opposite side is not the correct answer.

 

[dx]

You have to use the component of the worker's force in the direction motion of the box, i.e. parallel to the inclined plane. Always use the component of the force in the direction of the displacement when calculating work.

 

[Lothar]

So the 86N given is not parallel to the plane?

 

[dx]

What does it say in the question?

 

[Lothar]

Oh ok. The guy is pushing 86N parallel to the ground.
So is it 5m*86N*sin(36.87)?

 

[dx]

Nope. I suggest you draw a picture and mark the force vector and the angle it makes with the inclined plane.

 

[Lothar]

Got it. I used the compliment of the angle given.
I'll google the last part of the question. We haven't learned anything about coefficient of friction.
I'll stop bothering you now.
Thanks for the help.

 

[Lothar]

I don't understand it.
How do you use coefficient of friction?

 

[ImAnEngineer]

Friction = coefficient of friction * Normal force