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# Dynamics w/ Friction ## Question hi anyone out there, i need help with a problem. it is asking me which combination of three concurrent forces acting on a body could not produce equilibrium?

1) 1N, 3N, 5N 2) 2N, 2N, 2N 3) 3N, 4N, 5N 4) 4N, 4N, 5N

i thought i was doing it right, but the only answer that would produce equilibrium is 3) right? what are the other two? how would i go about doing a problem like this?

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• 0 well you know 3 could because if you put 3N and 4N at 90 degrees there resultant would be 5N, so if you put the 5N in the opposite direction of the resultant the object would not move.

2 could work too, because if you set it up so that the 3 2N forces were 120 degrees apart, a peice, you would be able to achieve an equilibrium (draw it!)

Now that i think about it, the answer is 1. 1 could not have an obtainable equilibrium. When you put the other ones at funky angles you can always achieve an equilibrium. BUT, for number 1, even if you had the 5N force going one way, and the 1N and 3N force ON TOP of eachother heading in the exact opposite direction, you could not add up to the 5N needed to attain an equilibrium. Adding an angle between the forces only reduces the resultant (there is no sinx or cosx value bigger then 1).

So to do this problem, the first step would be to see if 2 of the resultants can add up to be greater then or equal to the other resultant in all cases. For number 1, 1N + 3N < 5N, so it would be impossible to have an equilibrium

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• 0 sweet! thank you so much for your help. one last question, how does 4N, 4N, and 5N work then? how do i determine the angles necessary for equilibrium?

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You might have to play for a bit to determine the angles (the math gets a bit more involved), but it's pretty straightforward to see how you could line up a 4N, 4N, and a 5N vector to achieve equilibrium... I've used the vector simulation lab from PHET (http://phet.colorado.edu) to demonstrate how putting a 4N, 4N, and 5N vector together gives you a vector sum of 0.

[ATTACH=CONFIG]52[/ATTACH]

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• 0 thank you that visual helps!

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• 0 Can anyone help me with this problem?

A book weighing 20 N slides at a constant velocity down A ramp inclined 30 degrees to the horizontal. What is the force of friction between the book and the ramp?

Any sort of help would be greatly appreciated!

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• 0 Nevermind, Phoebe helped me ##### Share on other sites

• 0 Okay this time I am in dire need of help - A 50N horizontal force is needed to keep an object weighing 500 N moving at a constant velocity of 2 m/s across a horizontal surface. what is the magnitude of the frictional force acting on the object?

I have looked everywhere in my notes and cannot find any help ##### Share on other sites

• 0 If the object is at constant velocity (that means it is not accelerating) then Fnet= 0 since F=ma and a=0 at constant velocity. If it takes 50N to pull it at constant velocity, then the force of friction must be 50N in the opposite direction for Fnet to be 0!

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• 0 Wait then how is it that the object is moving 2 m/s if the forces are the same? If the forces are the same, shouldn't the object be motionless?

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• 0 not if the object accelerated until it got to 2 m/s but when it reached that point the force would have had to change to a point where it equaled the force of friction. It doesnt say anything about what happend before 2 m/s so you dont have to worry about that

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Great response... Moe is absolutely right.

Remember, no net force doesn't mean an object is at rest. no net force means an object won't accelerate -- therefore an object at rest will remain at rest, and an object in motion will remain in motion at a constant velocity and in a straight line -- unless there is a net force.

For example, if a spaceship is moving at 1000 m/s through space, if Superman pushes it forward with a force of 50N and The Hulk pushes it backward with a force of 50N, the forces are balanced, so there is no net force, therefore the spaceship continues at 1000 m/s. Net force = 0 implies no acceleration.

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• 0 P.S. Feel free to ask all the questions you can! I plan to major in engineering physics next year (in addition to electrical engineering) so it is helpful to review by figuring out old problems!

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If you weren't a super duper physics student, it would seem so! However, Forces cause accelerations on an object. and if an object is not accelerating in any direction, two things can be happening to the object. It can either be motionless OR in constant motion; talk to yourself about the definition of constant velocity, thats how I wrapped my brain around that one. Velocity only changes with acceleration (forces), so a constant velocity means no forces or forces that cancel each other out. I hope this helps!

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Wait then how is it that the object is moving 2 m/s if the forces are the same? If the forces are the same, shouldn't the object be motionless?

It would seem so. What you have to keep in mind Newton's Second Law of Physics F=ma. A force results only in an acceleration. AND we know that if an object has Fnet=0 (that means that there is either no forces acting on an object, or that the forces acting on the object cancel each other, as in this case) then the object is not accelerating in any direction. There are only two possibilities for an object that is experiencing no accelerations; this comes from Newton's First Law: an object will remain at rest or in constant motion unless acted upon by an outside force. Your object, however, also has a velocity. Ergo, your object is moving with constant velocity. Hope this helps!

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