# Newton's Second law problem

1. Apr 14, 2013

### Clever_name

1. The problem statement, all variables and given/known data

See the this link for problem statement, http://i48.tinypic.com/vphsm1.jpg

Case 1 I've solved, however I'm unsure about how good my solution is for case 2.

2. Relevant equations
F=ma Fu = N * uk

3. The attempt at a solution

Ok first I sum the forces in the y direction orientated about the log so my y direction is normal to the log.

N - 400(9.81)cos(a) = 0

then i do a force summation in the x direction.
0.5*N -2T +400*9.81sin(a)=0

next I do a force summation on block A in the upward direction, where the downward direction is positive,

125(9.81)-T = 0
so T =1226.25 N

Now this where i get stuck.
I have two equations with two unknowns but can't solve for a. my attempt is as follows:

N=400*9.81*cos(a)

then i sub this into 0.5*N -2T + 200*9.81sin(a)
to get 1962cos(a) -2452.5 + 3924sin(a) =0

but i cannot solve this equation for a.

Any tips on where I'm going wrong would be greatly appreciated. Thanks!

Last edited: Apr 15, 2013
2. Apr 14, 2013

### haruspex

It says 200kg, not 400.
= 0? The log is accelerating but the block isn't?

3. Apr 15, 2013

### Clever_name

Hi Haruspex,

I'm working on case 2

4. Apr 15, 2013

### haruspex

Sorry - missed that.
In case 2, the log is on the point of slipping down. Which way is the friction acting?
400
Your use of N for normal and N for Newtons is confusing.
Given a sin θ + b cos θ = c, divide both sides by √(a2+b2). Then find ψ s.t. a/√(a2+b2) = cos ψ, etc.

5. Apr 15, 2013

### Clever_name

You said "Given a sin θ + b cos θ = c, divide both sides by √(a2+b2). Then find ψ s.t. a/√(a2+b2) = cos ψ, etc."
I really am confused by the second sentence.... could you explain it to me i don't know what "s.t. a/√(a2+b2) = cos ψ" means.

Thanks!

6. Apr 15, 2013

### haruspex

s.t. is an abbreviation for "such that". You can find ψ from a/√(a2+b2) = cos ψ, right? It follows that sin ψ = b/√(a2+b2). So you have
cos ψ sin θ + sin ψ cos θ = c/√(a2+b2) = sin (θ+ψ). Solve that to find θ+ψ.

7. Apr 15, 2013

### Clever_name

Ok my attempt is as follows:

1962cos(∅) +3924sin(∅)= 2452.5

first i divide through by 3924, to get 0.5cos(∅)+sin(∅)=0.625.

so a/(a^2+b^2)^1/2= cos(ψ)

so 1/(1^2 +0.5^2)^1/2 =cos(ψ) so ψ = 26.57

then c/√(a2+b2) = sin (θ+ψ)

so 0.625/(1^2 +0.5^2)^1/2 = Sin(∅+26.57)

solving ∅= 7.42?

8. Apr 15, 2013

### haruspex

You tell me. Does that value satisfy 1962cos(∅) +3924sin(∅)= 2452.5?
But don't forget my question about which way friction is acting here.

9. Apr 15, 2013

### Clever_name

I don't understand, at the critical slope the log is not moving thus there is no kinetic friction?

10. Apr 15, 2013

### haruspex

I agree the question is a little odd here. It says to determine the max slope such that the log will not slide down when the block is released. That sure sounds like static friction to me, but in the preamble it discusses kinetic friction.
Either way, the point is that the log is in danger of slipping down, so presumably if the friction were any less it would do so. Which way, then, is the friction acting?

11. Apr 16, 2013

### Clever_name

opposite the direction of motion, so to the left.

12. Apr 16, 2013

### haruspex

The log is in danger of slipping down the slope. Does friction act to oppose relative motion of the surfaces or to assist it?

13. Apr 16, 2013

### Clever_name

friction acts to oppose relative motion of the surfaces - therefore assuming the log moves to the right frictional force will act to the left, however if the log is in danger of slipping down the slope, then frictional force will act to the right.

14. Apr 16, 2013

### Clever_name

however in the problem there is no mention of static friction.

15. Apr 16, 2013

### Clever_name

are you inferring that my solution is incorrect haruspex?

16. Apr 16, 2013

### haruspex

Yes. Let's set aside the kinetic v. static friction issue for the moment. You are looking for the max angle such that the log will not slip down. This means that without friction it would slip down, which is to the left. Therefore friction is acting up the slope. This makes your calculation wrong.
It would have been better if the preamble in the question had referred to static friction, not kinetic. But unless you assume they meant static there is no way to answer the question.

17. Apr 25, 2013

### Clever_name

Ok now I've assumed we are talking about static friction my final attempt is as follows:

2T+Mgcos(a)-Mgsin(a)=0

and T =1226.25

but solving the first equation i can't see how to go about this because the minus sign doesn't allow me to use that trig identity you kindly suggested earlier.

Any suggestions are appreciated.

nvm - i now see that i can use that trig identity.

Last edited: Apr 25, 2013
18. Apr 25, 2013

### Clever_name

haruspex don't you mean c/√(a2+b2) = cos (θ+ψ)? nvm i'm confusing myself.

Last edited: Apr 25, 2013
19. Apr 30, 2013

### Clever_name

Ok so this problem is still giving me some trouble. My current attempt reads as follows:

do a force balance at position c
2T + (Mc)*gcos(∅)*Us - (Mc)*gsin(∅) = 0

do a force balance at A,

T=1226.25

sub back into equation 1,

-3924sin(∅)+1962cos(∅)=-2452.2

then using a/√(a2+b2) = cos ψ

i get ψ = 153.43

then using c/√(a2+b2) = sin (θ+ψ)

i get -187.41 but this answer is nonsense :(

any tips on where i'm going wrong would be appreciated thanks!

20. Apr 30, 2013

### haruspex

Your value for ψ is correct, but sin(ψ) = b/√(a2+b2) (i.e. positive) so you have cos(ψ)sin(∅)+sin(ψ)cos(∅)=-2452.5/√(a2+b2) = cos(ψ-∅) = cos(124 deg).