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1) What force is necessary to drag a 20N box up a 45 degrees incline plane if the coefficient of friction is .47?

I think I understand some of this. Ff= Coefficient of Friction x Fn.
Coefficient of Friction: .47
W: 20N
Fn=W times Cos)Theta.
Fn=20N times Cos)45= 14N
Ff= Coefficient of Friction x Fn
Ff= .46 x 14N= 6.6N

This is where I got lost. Any help please?

2) The coeffient of a Sliding friction between a crate and a horizontal floor is .27. If the crate has a mass of 250kg, what force is needed to slide it across the floor at constant velocity?

I think I got this one.
Coefficient of Friction= .27
M: 250kg
W: 250kg x Gravity(9.81)= 2453N
Ff= Coefficient of Friction x W
Ff=2453N x .27= 662N

Thanks alot for those who helps.

2007-11-12 04:43:56 · 3 answers · asked by Lying from you 4 in Science & Mathematics Physics

3 answers

1) Look at a FBD of the box

Assume a constant speed
F-f-W*sin(th)=0
you are right that
f=Fn*0.47
F=6.65+20*sin(th)
F=20.8 N

2) Looks correct

j

2007-11-12 05:03:23 · answer #1 · answered by odu83 7 · 0 0

1) To pull the box up the plane, you've got to do two things. First, you have to overcome the friction to make the box slide. You've worked out the friction. You figured the component of the box's weight which acts into the plane.
Second, you have to lift the box up - increase it's height. Imagine there was no friction, and resolve the component of the weight of the box which acts down the incline. Once you've overcome the firction, you still have to apply that force too. Now just add a tiny, tiny bit of force, and the box will start to move up the plane.
Once you've done all your workings (use at least 2 decimal places), you should get an answer of 20.79N

You have the second part right

2007-11-12 05:04:16 · answer #2 · answered by Experimentor 2 · 0 0

that's labeled as a dynamics difficulty. a million. The impact is comparable to the launch velosity. by way of fact the preliminary velosity replaced into 8 m/s its impact velosity additionally = 8 m/s 2. finished flight time = (2 x preliminary velosity x Sin of perspective)/g finished time = (2 x 8 m/s x Sin ninety ranges) /9.80 one m^2 = a million.sixty 3 sec

2016-12-08 19:41:07 · answer #3 · answered by lirette 4 · 0 0

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