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Posted

Three questions.

 

1) A skier has just began descending down a 20 degrees slope. Assuming that the coefficient of kinetic friction is 0.10, calculate the acceleration of the skier and the final velocity after 8 sec.

 

We don't have enough information to solve this...or am I missing something?

 

2) A traffic light with a mass of 30kg is hung in equilibrium by 2 wires, one at an angle of 45 degrees, the other at an angle of 37 degrees. Find the tension in the two wires.

 

3) One bucket weighing 25N is hanging by a massless cord from another 25N bucket, which in turn is hanging from a third 25N bucket. All three are being lowered with an acceleration of 1.5m/s^2. Calculate the tension in each cord.

Posted

Ok, ignore the last two questions.I hadn't realized that the tension was the opposite of the force of gravity (and then calculate in necessary angles and acceleration...)

 

So for the first question, we have the coefficient of 0.10 mu. I know we can find F net from that, but we don't have anything else.

 

Can;t upload picture... =(

Posted

Draw your picture and include the broken down vectors. You know that FFriction=µN and that Fnet=ma.

 

What have you tried thusfar with that info?

Posted

We have insufficient information. Without a mass, we can't calculate the force of anything.

 

And I would draw a picture (online) if I could...but it'll only let me attach URLs?

Posted
We have insufficient information. Without a mass, we can't calculate the force of anything.
Are you sure about that? Try the problem without putting numbers in and see if the mass cancels.

 

[hide]Hint: it does[/hide]

Posted

You have three questions about velocity, force and acceleration. Can you give me the formula that connects those? (Two formulas, ideally).

 

I ask, because once you look at the formulas, you know what to look for in the question. The next step would be to draw the images and write down all forces that apply.

 

And the important thing is *ALWAYS*: read carefully! ;)

 

Three questions.

 

1) A skier has just began descending down a 20 degrees slope. Assuming that the coefficient of kinetic friction is 0.10, calculate the acceleration of the skier and the final velocity after 8 sec.

You have all the information you need. Don't forget that the friction force isn't exactly opposite to the gravity. Draw it out... the gravitational force is downwards. Where is the friction?

 

 

 

 

 

 

Physics questions are usually straight forward, and going by formulas (at this stage, at least). Lay out the relevant formulas, draw out ALL the forces and look at the picture, see what you can get from all of it and if you really miss anything.

 

How did you start the first question? Where do you get stuck?

 

~moo

Posted

Th two equation are:

F net = m*a

F net = m* ([v2 - v2]/t)

 

There is also F friction = mu*F normal

 

We know:

mu = 0.10

v1 = 0

t = 8

a [down] = 9.8

an angle = 20 degrees

 

It's possible to find the acceleration of the object going down the slope. That would be

a (angled) = 9.8 sin20.

But unless I'm somehow really blind and am missing out on something (probably a mass or force), then I'm not sure where to go next.

 

My pathetic attempt of sketching it out is in my avatar.

Posted

Fnet=ma

 

=Fgrav+Ffric

 

=-m*sin(20)g+µN

 

=-m*sin(20)g+µm*sin(20)g

 

=mg*sin(20)(-1+µ)

 

ma=mg*sin(20)(-1+µ) => a=g*sin(20)(µ-1)

Posted

I broke it down more. Gravity is due to mass(and, more importantly, has a constant acceleration near the earth's surface) and all of the forces in this problem are from gravity. When the net force is some multiple of the gravitational force, you don't need to know the mass to find the acceleration.

 

I also broke the math apart more in my previous post. The rest of the question is simply applying the kinematic equations.

Posted

Well, that DOES make it easier for the eyes =P

 

Just one thing I'm not sure about the calculation. When you were grouping it at:

 

=-m*sin(20)g+µm*sin(20)g

 

=mg*sin(20)(-1+µ)

 

Wher did the -1 come from? Wouldn't it remain poistive?

 

And how is (m*sin(20)g) the F grav? Wouldn't that just be the fricitonal force?

Posted
Well, that DOES make it easier for the eyes =P

 

Just one thing I'm not sure about the calculation. When you were grouping it at:

 

=-m*sin(20)g+µm*sin(20)g

 

=mg*sin(20)(-1+µ)

 

Wher did the -1 come from? Wouldn't it remain poistive?

In your co-ordinate system(I'm using the one from your avatar), gravitational force is negative.

And how is (m*sin(20)g) the F grav? Wouldn't that just be the fricitonal force?

It's the amount of gravitational force in the direction of travel. Maybe you do use the whole thing, though. It's been a few years since I took physics.

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