Normal Force

[adrakdavra]

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Back to Jorge and his crate… Jorge pushes on a 10-kg crate 5 m with a 50 N force at 45 degrees. The coefficient of kinetic friction is 0.2. What is the total work?

How much work is done by friction in the Jorge/crate example? How much work is done by the normal force?
I push on a wall. Does the normal force do work? Why am I tired?
I push on a wall while on a skateboard, does the normal force do work?
Does the normal force ever do work?

 

[milski]

The normal force is always perpendicular to the direction of movement and never does work.

For the rest: you need to be a bit more clear: what is 45 degrees? Is he pushing a cart up on a ramp? Or a cart on flat ground with a force at 45 degrees with the ramp? Pointed up or down?

Draw a free body diagram, calculate the force parallel to the movement surface and multiply that by the distance of the movement along that surface.

 

[adrakdavra]

The normal force is always perpendicular to the direction of movement and never does work.

For the rest: you need to be a bit more clear: what is 45 degrees? Is he pushing a cart up on a ramp? Or a cart on flat ground with a force at 45 degrees with the ramp? Pointed up or down?

Draw a free body diagram, calculate the force parallel to the movement surface and multiply that by the distance of the movement along that surface.

cart on flat ground with a force at 45 degrees with the ramp? Pointed up

 

[milski]

cart on flat ground with a force at 45 degrees with the ramp? Pointed up

The horizontal component of the force is Sqrt(2)/2*50N=0.7*50N=35N.

Work done by Jorge: 5m*35N=175J

Friction force = -μN=0.2*sqrt(2)/2*mg=-0.2*0.7*10*10=-2*7=-14J

 

[PhysMatMan]

The horizontal component of the force is Sqrt(2)/2*50N=0.7*50N=35N.

Work done by Jorge: 5m*35N=175J

Friction force = -μN=0.2*sqrt(2)/2*mg=-0.2*0.7*10*10=-2*7=-14J

Friction=-μN but N does not equal sqrt(2)/2*mg in general. You have to add Jorge's push to gravity. N=mg-sqrt(2)/2*50

 

[milski]

Friction=-μN but N does not equal sqrt(2)/2*mg in general. You have to add Jorge's push to gravity. N=mg-sqrt(2)/2*50

Yes, you're right, I butchered N completely.

N=mg-sqrt(2)/2*50=100N-0.7*50=100-35=65N

Work by the friction force: 0.2*65N*5m=65J

 

[Nickmiller]

I was wondering if I could hijack the thread with another forces question?
If I have, for example, three forces acting on a box, one pointing in one direction (F1) and the other two pointing in the opposite direction (F2 and F3), and the object is moving in constant velocity (so acceleration=0), which of the following would you say is correct:

F1= F2+F3

or

F1+F2+F3=0

I thought the second one is correct but the answer states that the first one is correct. Do you not consider the direction in this case, only magnitude? Sorry if this is a silly question, it's just been bothering me. Any input would be greatly appreciated.

 

[milski]

I was wondering if I could hijack the thread with another forces question?
If I have, for example, three forces acting on a box, one pointing in one direction (F1) and the other two pointing in the opposite direction (F2 and F3), and the object is moving in constant velocity (so acceleration=0), which of the following would you say is correct:

F1= F2+F3

or

F1+F2+F3=0

I thought the second one is correct but the answer states that the first one is correct. Do you not consider the direction in this case, only magnitude? Sorry if this is a silly question, it's just been bothering me. Any input would be greatly appreciated.

The first is valid for the magnitudes, the second if consider the forces to have a sign. The first will get you positive numbers for all three, the second will have some of the forces with a minus sign. I prefer the second way, since you don't have to think about direction explicitly but if you know that you're working with magnitudes, the first one will do ok too.

 

[Nickmiller]

The first is valid for the magnitudes, the second if consider the forces to have a sign. The first will get you positive numbers for all three, the second will have some of the forces with a minus sign. I prefer the second way, since you don't have to think about direction explicitly but if you know that you're working with magnitudes, the first one will do ok too.

Thank you for the clarification