Object exerts an opposing force of equal magnitude

[Ukitake Jyuushirou]

Newton's 3rd law stated with every force u exert on an object, the object exerts an opposing force of equal magnitude. if so why is that we're able to move objects around?

 

[arildno]

Because the forces act on DIFFERENT objects.

 

[andrevdh]

Lets consider two cases. In the one case you push a box over the floor and in the other case you push against a wall. Obviously the answer lies in how hard the object can push back at you. If it has not moved when you reached your maximum strength, then you are in trouble, no motion of the object = no motion of you. The answer therefore lies in the system that you are considering (you + box). You need to look at the total force that the combined system are experiencing for motion (according to N2). Remember that the box will match your pushing force against it, but you are also pushing against the floor with your feet, and the floor are pushing back at you with the same force (N3 again). If this force overpowers, or matches, the matching force from the box, then we've got motion of the system. Note that you can decide how hard you want to push the system (you + box) by how hard your legs are pushing against the floor, thereby determining how quickly the system will move.

 

[Ukitake Jyuushirou]

Lets consider two cases. In the one case you push a box over the floor and in the other case you push against a wall. Obviously the answer lies in how hard the object can push back at you. If it has not moved when you reached your maximum strength, then you are in trouble, no motion of the object = no motion of you. The answer therefore lies in the system that you are considering (you + box). You need to look at the total force that the combined system are experiencing for motion (according to N2). Remember that the box will match your pushing force against it, but you are also pushing against the floor with your feet, and the floor are pushing back at you with the same force (N3 again). If this force overpowers, or matches, the matching force from the box, then we've got motion of the system. Note that you can decide how hard you want to push the system (you + box) by how hard your legs are pushing against the floor, thereby determining how quickly the system will move.

wouldnt the force N3 generated from the feet pushing against the floor be opposed by the box?

 

[HallsofIvy]

No, the box can't "apply" more force than the friction force on it. If you subtract the friction force from the force N3 you are applying to the box, the result is the "net" force that causes acceleration. Once again, when you press against the box with force N3, you feel the box pressing back against your hands with force N3 but part of that is friction force, the rest ma where a is the acceleration of the box. Your hands pushing on the box and the box are the "equal and opposite" action and reaction, but part of that reaction is due to the boxes acceleration.

 

[Ukitake Jyuushirou]

so when i push a box with my hands , the box also exerts an equal and opposing force but the reason why the box will move is becoz my feet is also exerting a force downwards on the Earth that pushes me back and that force is transmitted to the box and over comes the friction and becomes the net force...am i rite?

while we're at it...

can i find the coefficient of friction if i am given the normal force and the force that is directed horizontally on the object?

assuming normal force is 236N and the force directed horizontally is 69.2 N

 

[arunbg]

so when i push a box with my hands , the box also exerts an equal and opposing force but the reason why the box will move is becoz my feet is also exerting a force downwards on the Earth that pushes me back and that force is transmitted to the box and over comes the friction and becomes the net force...am i rite?

If force is to be maintained, you need to certainly use the friction to prevent you from going backward, while pushing the box forward.

can i find the coefficient of friction if i am given the normal force and the force that is directed horizontally on the object?

If you are talking about the coefficient of static friction, then the force directed horizontally must be the force applied just before the object starts to slide.
Use F=\mu_sN

If you are talking about coefficient of kinetic friction, then you also need to take acceleration of the mass into consideration.

___________________

Arun

Two things are infinite: the universe and human stupidity; and I'm not sure about the universe.
~ Albert Einstein

 

[arildno]

No. The box moves forward because there is a net force acting upon IT.
The box couldn't care less what force acts upon your feet, or if indeed, you're wearing sneakers.

 

[Doc Al]

wouldnt the force N3 generated from the feet pushing against the floor be opposed by the box?

"N3" is andrevdh's shorthand for "Newton's 3rd Law"; it's not identifying a specific force.

If the feet push against the floor, then the floor pushes back against the feet (not on the box directly). That's what N3 tells you. (Of course, the floor pushing against your feet allows you to push harder against the box. Lot's of luck trying to push the box if you are standing on ice.)

 

[andrevdh]

Lets consider another case. Two identical blocks resting on a frictionless surface. You push the block closest to you with your finger resulting in the blocks moving forwards together. The resultant force that the blocks are experiencing comes from your finger pushing them. The action-reaction forces in between the two blocks are equal and are not contributing to the motion of the blocks. They are called internal forces. The same can be said regarding the action-reaction pair of forces in your question. In the case of the two blocks the force coming from your finger is like the force pushing against your feet in my other example.

 

[andrevdh]

wouldnt the force N3 generated from the feet pushing against the floor be opposed by the box?

N3 refers to Newton's third law. When you push the box it pushes back against you with a force of same magnitude (according to N3). Your feet pushing against the floor will generate another force by the floor pushing against you (N3 again). The reaction force from the box and the floor are not the same forces, they can differ in magnitude depending on how hard you push against the floor. The example in my previous post about two similar blocks clarifies the situation better.

You can see that the force from the finger and the force that the front block experiences will be different in magnitude using N2 in the situation. The force from your finger have to accelerate both blocks, while the interaction force that the block in the front experiences just have to accelerate that block alone.

 

[Ukitake Jyuushirou]

erm pardon my poor understanding of physics

so as long as the net force generated cancels the friction force, the box will move. as with all pushing motion a force is directed at the box and another force from my feet generated downwards as long as they overcome friction force, the box will move?...

 

[Doc Al]

Keep it simple: As long as the net force on the box is nonzero, the box will accelerate. If your (horizontal) pushing force on the box exceeds the static friction force of the floor on the box, then there will be a net force on the box and it will move.

 

[Ukitake Jyuushirou]

ok i'd ignore the messier bit regarding the force generated from the feet. thanks :)

 

[Ukitake Jyuushirou]

If you are talking about the coefficient of static friction, then the force directed horizontally must be the force applied just before the object starts to slide.
Use F=\mu_sN

If you are talking about coefficient of kinetic friction, then you also need to take acceleration of the mass into consideration.

once the force applied to the box is greater than the static friction, i can safely ignore the static friction force right? since at this point only the kinetic friction comes into play, but if given the 2 forces and i am told to find co efficient of kinetic friction, how will i do so?

 

[Doc Al]

I'm not sure what you are asking. If the box is sitting still and you push it, static friction will match your force to keep the box from sliding--up to the point where the static friction is at its maximum value, which is \mu_s N. As you state, once it starts moving, kinetic friction takes over; that force equals \mu_k N.

How you would find \mu_k depends on what you know. If you know the acceleration, you can apply Newton's 2nd law to find the friction force, and from that deduce \mu_k.

 

[Ukitake Jyuushirou]

I'm not sure what you are asking. If the box is sitting still and you push it, static friction will match your force to keep the box from sliding--up to the point where the static friction is at its maximum value, which is \mu_s N. As you state, once it starts moving, kinetic friction takes over; that force equals \mu_k N.

How you would find \mu_k depends on what you know. If you know the acceleration, you can apply Newton's 2nd law to find the friction force, and from that deduce \mu_k.

the question is actually: a 20kg sled is pulled along a horizontal surface at constant velocity. The pulling force has a magnitude of 80N and is directed at an angle of 30 degrees above the horizontal, determine the coefficient of kinetic friction

using kinematics i find the force in the positive y-component is 236N and the force of gravity is 196N. a horizontal force along the positive x-component is 69.2N , here is where i got lost

 

[Doc Al]

Let's do it step by step. First identify all the forces acting on the sled. I count four:
(1) Pulling force (what are the x & y components?)
(2) weight (what are the x & y components?)
(3) normal force (what are the x & y components?)
(4) kinetic friction (what are the x & y components?)

Note: The direction of the forces (and thus the sign of the component) counts.

 

[Ukitake Jyuushirou]

Let's do it step by step. First identify all the forces acting on the sled. I count four:
(1) Pulling force (what are the x & y components?)
(2) weight (what are the x & y components?)
(3) normal force (what are the x & y components?)
(4) kinetic friction (what are the x & y components?)

Note: The direction of the forces (and thus the sign of the component) counts.

pulling force

x = +69.2
y = +40

weight = -196

normal = +196

kinetic friction = ??

 

[Doc Al]

pulling force

x = +69.2
y = +40

OK

weight = -196

OK (what component?)

normal = +196

How did you get this? (Apply equilibrium conditions.)

kinetic friction = ??

Apply equilibrium conditions.

The first two forces are given; the last two you solve for by applying the condition for equilibrium: The sum of the forces in any direction must be zero.

 

[arunbg]

normal = +196

No, note that the pull and the weight of the object act in different directions. So what is the apparent weight of the sled as felt by the ground ?
What is the expression for kinetic frictional force ? Are the forces in the horizontal direction balanced ?

 

[Ukitake Jyuushirou]

OK


OK (what component?)


How did you get this? (Apply equilibrium conditions.)


Apply equilibrium conditions.

The first two forces are given; the last two you solve for by applying the condition for equilibrium: The sum of the forces in any direction must be zero.

weight = 196 (negative y-component)

normal force is juz the opposite of weight isn't it?

 

[HallsofIvy]

"Constant velocity"= "accleration 0". Since the sled moving with 0acceleration, the net force, 40N- kinetic friction= 0. That should make it easy to find the kinetic friction!
Since kinetic friction= (coefficient of friction)*(weight),
coefficient of friction= kinetic friction/weight.

 

[Doc Al]

normal force is juz the opposite of weight isn't it?

No. That's only the case if the applied forces have no vertical component. Read arunbg's comment.

In any case, the normal force is a passive force: you don't assume what it is, you figure it out. Apply this: The net force in the vertical direction must equal zero. (Add up the vertical components of all four forces and set that equal to zero.) That will tell you what the normal force is.

 

[Doc Al]

Since kinetic friction= (coefficient of friction)*(weight)

You mean normal force, which in this case is less than the weight.

 

[Ukitake Jyuushirou]

here is how i percieve the diagram...i hope its right

 

[Doc Al]

here is how i percieve the diagram...i hope its right

You left off the friction force and you presumed (incorrectly) that the normal force equals the weight.

If I were drawing that diagram, I would show:
-The pulling force (since it's given) and its x & y components
-The weight, which is given
-The normal force, which is unknown so mark it as Fn. (You should know its direction, of course.)
-The friction, which is unknown so mark it as Ff. (You should know its direction, of course.)

 

[Ukitake Jyuushirou]

updated...

 

[Doc Al]

Much better. Now solve for Fn and Ff. (Show your equations.)

 

[Ukitake Jyuushirou]

since the weight of the sled is - 196N downwards and if there are no vertical forces, the Fn should be + 196 N but since the pull supplies a 40 N positive force, then the Fn will be 156

correct?

 

[arildno]

since the weight of the sled is - 196N downwards and if there are no vertical forces, the Fn should be + 196 N but since the pull supplies a 40 N positive force, then the Fn will be 156

correct?

For your own sake:
Gain familiarity and practice with equation solving by doing this with standard mathematical formalism!

 

[arunbg]

I think he will learn mathematical formalism in due time

Yes, your value for normal force is correct .

 

[Ukitake Jyuushirou]

For your own sake:
Gain familiarity and practice with equation solving by doing this with standard mathematical formalism!

wat is mathematical formalism? pardon me as i have not been doing physics in a long time and the last time i did physics was really simple and rudimentary physics. nothing beyond blind application of F = ma and the likes...

i'm pretty much trying to figure out how to study physics correctly

 

[Doc Al]

since the weight of the sled is - 196N downwards and if there are no vertical forces, the Fn should be + 196 N but since the pull supplies a 40 N positive force, then the Fn will be 156

correct?

Yes, but please do not try to do these calculations in your head! Do as arildno (and I) suggests: Write the equation for vertical equilibrium. Then solve the equation.

When you've done it formally (and correctly) a zillion times, then you can take shortcuts.

 

[Ukitake Jyuushirou]

Yes, but please do not try to do these calculations in your head! Do as arildno (and I) suggests: Write the equation for vertical equilibrium. Then solve the equation.

When you've done it formally (and correctly) a zillion times, then you can take shortcuts.

ok got it

w=mg
196=20x9.8

Fn = 196 (no vertical force)

196-40 = 156 (force of 40N in +y-component)
Fn = 156

 

[Ukitake Jyuushirou]

now i have to solve for the coefficient of kinetic friction. kinetic friction force is supposed to be proportional to the Fn. but from wat is given, how do i work out the friction force?

 

[arunbg]

What is the constant of proportionality in the equation relating kinetic frictional force and normal force ?
What are the conditions for horizontal equilibrium ?
As Doc Al suggested, write down the math step by step .

 

[Ukitake Jyuushirou]

frictional force is related to the Fn.

Fn = 156, Fx = 69.2. coefficient of friction is 69.2 divided by 156 = 0.44

am i correct?

 

[Doc Al]

w=mg
196=20x9.8

Fn = 196 (no vertical force)

196-40 = 156 (force of 40N in +y-component)
Fn = 156

Good. But just for the record, here's how I would do it:
(Sum of the forces in the vertical direction) = +Fn -mg +80sin(30) = 0

That's the equation for vertical equilibrium. Solving it (trivial, yes) gives:
Fn = mg - 80sin(30)

Now do the same for horizontal forces.

 

[Ukitake Jyuushirou]

Good. But just for the record, here's how I would do it:
(Sum of the forces in the vertical direction) = +Fn -mg +80sin(30) = 0

That's the equation for vertical equilibrium. Solving it (trivial, yes) gives:
Fn = mg - 80sin(30)

Now do the same for horizontal forces.

using ur formula


Fx = 80cos30 = 69.2

 

[Doc Al]

OK. Again, the way I would do it is:
(Sum of the forces in the horizontal direction) = -Ff +80cos(30) = 0

Solving gives: Ff = 80cos(30)

(The reason for sticking to a systematic approach is: (1) It drives home the key physics principles & (2) It allows you to solve much harder problems.)

Now you can apply the other relationship you know about kinetic friction to solve for \mu_k systematically.

 

[Ukitake Jyuushirou]

OK. Again, the way I would do it is:
(Sum of the forces in the horizontal direction) = -Ff +80cos(30) = 0

Solving gives: Ff = 80cos(30)

(The reason for sticking to a systematic approach is: (1) It drives home the key physics principles & (2) It allows you to solve much harder problems.)

Now you can apply the other relationship you know about kinetic friction to solve for \mu_k systematically.

yes ur right. i'd try to use this approach for my other qns

frictional force is related to the Fn.

Fn = 156, Fx = 69.2. coefficient of friction is 69.2 divided by 156 = 0.44

am i correct?

 

[Doc Al]

Looks good. I'd write it this way. The key relationship for kinetic friction is:
F_f = \mu_k F_n

thus:
\mu_k = F_f /F_n

then just plug in the numbers

 

[Ukitake Jyuushirou]

whew...one question down...many many more to go... thanks for everyone's help and advice

 

[Doc Al]

I think we beat that one to death!

 

[arunbg]

We sure did