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Ukitake Jyuushirou
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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?
wouldnt the force N3 generated from the feet pushing against the floor be opposed by the box?andrevdh said: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.
If force is to be maintained, you need to certainly use the friction to prevent you from going backward, while pushing the box forward.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 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.can i find the coefficient of friction if i am given the normal force and the force that is directed horizontally on the object?
"N3" is andrevdh's shorthand for "Newton's 3rd Law"; it's not identifying a specific force.Ukitake Jyuushirou said:wouldnt the force N3 generated from the feet pushing against the floor be opposed by the box?
Ukitake Jyuushirou said:wouldnt the force N3 generated from the feet pushing against the floor be opposed by the box?
arunbg said: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 [itex]F=\mu_sN[/itex]
If you are talking about coefficient of kinetic friction, then you also need to take acceleration of the mass into consideration.
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 frictionDoc Al said: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 [itex]\mu_s N[/itex]. As you state, once it starts moving, kinetic friction takes over; that force equals [itex]\mu_k N[/itex].
How you would find [itex]\mu_k[/itex] 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 [itex]\mu_k[/itex].
pulling forceDoc Al said: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.
OKUkitake Jyuushirou said:pulling force
x = +69.2
y = +40
OK (what component?)weight = -196
How did you get this? (Apply equilibrium conditions.)normal = +196
Apply equilibrium conditions.kinetic friction = ??
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 ?normal = +196
weight = 196 (negative y-component)Doc Al said: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.
No. That's only the case if the applied forces have no vertical component. Read arunbg's comment.Ukitake Jyuushirou said:normal force is juz the opposite of weight isn't it?
You mean normal force, which in this case is less than the weight.HallsofIvy said:Since kinetic friction= (coefficient of friction)*(weight)
You left off the friction force and you presumed (incorrectly) that the normal force equals the weight.Ukitake Jyuushirou said:here is how i percieve the diagram...i hope its right
For your own sake:Ukitake Jyuushirou said: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?
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...arildno said:For your own sake:
Gain familiarity and practice with equation solving by doing this with standard mathematical formalism!
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.Ukitake Jyuushirou said: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?
ok got itDoc Al said: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.
This means that when one object pushes or pulls on another object, the second object will push or pull back with the same amount of force. This is known as Newton's third law of motion.
This is due to the conservation of momentum. When two objects interact, they exchange momentum and the total amount of momentum in the system remains constant. Therefore, the object that is being pushed or pulled will exert an equal and opposite force to maintain this balance.
The magnitude of the opposing force depends on the amount of force applied by the first object. The two forces will always be equal in magnitude, but in opposite directions.
No, according to Newton's third law, the two forces must be equal in magnitude and opposite in direction. If one object exerts a larger force, the other object will exert an equal and opposite force to maintain balance.
No, the magnitude of the opposing force is solely dependent on the amount of force applied by the first object. The type or size of the objects does not have an impact on this principle.