You are Here: Home

# Mechanics Help! watch

1. (Original post by 2776)
And so since F=uR

We know F and we know R.

and so u=F/R

and I get u = 0.349
Which does not correspond with the u that we know already
2. (Original post by 2776)
And so since F=uR

We know F and we know R.

and so u=F/R

and I get u = 0.349
hmmmm.....i still think μ should be constant....0.364.
3. (Original post by king of swords)
hmmmm.....i still think μ should be constant....0.364.
If yours comes out thinking that μ is less i guess that is a good thing because it really means that F has not reached its highest value yet.
4. (Original post by king of swords)
hmmmm.....i still think μ should be constant....0.364.
Yes, we need to prove that u is constant.
5. (Original post by 2776)
Yes, we need to prove that u is constant.
No you don't.....μ just IS constant....it's that being constant that you should use to prove the system is in equilibrium.
6. (Original post by king of swords)
No you don't.....μ just IS constant....it's that being constant that you should use to prove the system is in equilibrium.
so what else should I do to my calculation?
7. (Original post by 2776)
so what else should I do to my calculation?
well i'll just show you my confusion of it all:
if in equilibrium:
F + 250gsin20 = 2000cos20
R + 2000sin20 = 250gcos20
therefore FMAX = μR = μ(250gcos20 - 2000sin20) = tan20(250gcos20 - 2000sin20) = 589.0N

Also F = 2000cos20 - 250gsin20 = 1041.4N!!!! which exceeds FMAX...therefore i cannot see how this can be in equilibrium
8. (Original post by king of swords)
well i'll just show you my confusion of it all:
if in equilibrium:
F + 250gsin20 = 2000cos20
R + 2000sin20 = 250gcos20
therefore FMAX = μR = μ(250gcos20 - 2000sin20) = tan20(250gcos20 - 2000sin20) = 589.0N

Also F = 2000cos20 - 250gsin20 = 1041.4N!!!! which exceeds FMAX...therefore i cannot see how this can be in equilibrium
I think your 2 equations are wrong. See me 2 equations (latest one) on page 2. That is the ocrrect one
9. R- (250gcos20+2000sin20)=0

F+ (2000cos20-250gsin20)=0
10. (Original post by 2776)
R- (250gcos20+2000sin20)=0

F+ (2000cos20-250gsin20)=0
I'm guessing you have F going up the plane? and yes i've messed up the equation in R.
11. (Original post by king of swords)
I'm guessing you have F going up the plane? and yes i've messed up the equation in R.
That is correct. But then it is totally wrong. Cause I just checked, I get u= -0.348.

Which is totally wrong.
12. Cause F= 2986.287...

R= -1041.4...
13. (Original post by 2776)
That is correct. But then it is totally wrong. Cause I just checked, I get u= -0.348.

Which is totally wrong.
Well F should be going down the plane if the horizontal 2000N force is going "into" the plane. Remember, F changes to work against motion.
14. (Original post by king of swords)
Well F should be going down the plane if the horizontal 2000N force is going "into" the plane. Remember, F changes to work against motion.
done it....wait a minute
15. (Original post by king of swords)
Well F should be going down the plane if the horizontal 2000N force is going "into" the plane. Remember, F changes to work against motion.
Thats clever, yes I get it now. As 2000N is used to push up the plane, F should oppose action.

I mean this question is a killer isnt it? Its in OCR 2001 test. Did u do that?
16. (Original post by king of swords)
done it....wait a minute
F does go down the plane:
if in equilibrium:
F = 2000cos20 - 250gsin20 = 1041.4N
R = 2000sin20 + 250gcos20
so FMAX = μR = tan20(2000sin20 + 250gcos20) = 1086.9N which means that FMAX is greater than the actual value of F in this circumstance which means the system is in equilibrium.....sorry about that.
17. (Original post by king of swords)
F does go down the plane:
if in equilibrium:
F = 2000cos20 - 250gsin20 = 1041.4N
R = 2000sin20 + 250gcos20
so FMAX = μR = tan20(2000sin20 + 250gcos20) = 1086.9N which means that FMAX is greater than the actual value of F in this circumstance which means the system is in equilibrium.....sorry about that.
Right another question I'm stuck on:

particle A: 0.15kg (2m/s towards B)
particle B: 0.2 kg (stationary)

The coeffient of friction between A and surface is 0.05. a is brought to rest after collision. What is speed of B?

If it was a smooth surface, I have calculated that B moves at 3m/s away.
18. (Original post by 2776)
Right another question I'm stuck on:

particle A: 0.15kg (2m/s towards B)
particle B: 0.2 kg (stationary)

The coeffient of friction between A and surface is 0.05. a is brought to rest after collision. What is speed of B?

If it was a smooth surface, I have calculated that B moves at 3m/s away.
The particles are 4 m away form each other. I reckon that I should find the ending velocity i.e. u (2).
19. (Original post by 2776)
Right another question I'm stuck on:

particle A: 0.15kg (2m/s towards B)
particle B: 0.2 kg (stationary)

The coeffient of friction between A and surface is 0.05. a is brought to rest after collision. What is speed of B?

If it was a smooth surface, I have calculated that B moves at 3m/s away.
It would help if you knew whether the collision was inelastic or elastic.
20. (Original post by king of swords)
It would help if you knew whether the collision was inelastic or elastic.
Im not too sure, I think it wants me to use the momentum thing.

Anyway, the speed of B is 1.5m/s after collision on a smooth plane(not 3m/s)

Doesnt give the info, whether it is selastic or not.

TSR Support Team

We have a brilliant team of more than 60 Support Team members looking after discussions on The Student Room, helping to make it a fun, safe and useful place to hang out.

This forum is supported by:
Updated: December 7, 2003
Today on TSR

### He lied about his age

Thought he was 19... really he's 14

Poll
Useful resources

## Articles:

Debate and current affairs forum guidelines