# Homework Help: Newton, Find T

1. Oct 6, 2012

### SherBear

It's the same diag. but the blocks are A and B and the angle is 30°

Theta=30°
Block A=15kg
Block B=12kg
assume the rope is massless
us=.25=coefficient of static friction
What is T?

My Attempt at a solution
I made FBD for both block A and B

For A sum the forces
Fx=t1-m1gsin30°-f=ma
fy=N1-m1gcos30°=0

Block B
F=m2g-T2=m2a

So I'm looking for the Tension in the rope. Is it just 1 rope or 2 ropes? I really don't know what to do now. Do I find t2 before i find t1? Do I find the acceleration before I find the Tension?

Thank you!

2. Oct 6, 2012

### tiny-tim

Hi SherBear!

(try using the X2 button just above the Reply box )
from the diagram, it's one continuous rope, so (if the pulley is smooth) the tension is the same all the way along, and you can just call it T
fine so far

now write f in terms of N, and you'll have two equations with two unknowns (a and T) …

eliminate one and find the other

3. Oct 6, 2012

### SherBear

In terms of N, Would that be.... n1=m1gcos30° ?

4. Oct 6, 2012

### tiny-tim

no, f = µN1

(btw, did you get an email notification of this answer? i didn't get one for your answer, and only noticed it by accident some members are complaining that the notifications aren't operating properly)

5. Oct 6, 2012

6. Oct 6, 2012

### SherBear

u=.25 and n1 is? I have 2 unknowns, f and n1? I am sorry I'm really not that smart

7. Oct 6, 2012

### SherBear

My Professor says solve for N then plug into the Fx=T-mgsin20-f=ma or is it -ma, ?

8. Oct 6, 2012

### tiny-tim

(hmm … i got both of those notifications! )
yes, f = .25 n1 is the extra equation

(except, i've just noticed, that's the coefficient of static friction, so it only applies if a = 0 … perhaps he means dynamic friction?)

(and yes n1 is one of your unknowns, but you can get it from your n1 - m1gcos30° = 0 )

(btw, best to stick capital letters for forces, small letters for distances speeds accelerations etc)
depends which way you're measuring a …

you must measure a for both masses in the same direction (along the rope)

9. Oct 6, 2012

### SherBear

usf is the only coefficient for friction he gave us so for Fx Fy and F for B all = 0 in the equations if it is static friction?

using N1-m1gcos30=0 i got
m1gcos30=N1?
(15kg)(9.80m/s^2)(cos30)=N1
127.30N=n1

then use f=uN1
f=(.25)(127.30N)
f=31.825N?

10. Oct 6, 2012

### tiny-tim

looks ok

11. Oct 6, 2012

### SherBear

Thank you for your time and patience Tim. Is the final answer, f=31.825N? the Tension or?

12. Oct 6, 2012

### tiny-tim

uhh?

f is the friction

13. Oct 6, 2012

### SherBear

How do I find tension? I really don't know what to do next?

14. Oct 6, 2012

### tiny-tim

15. Oct 6, 2012

### SherBear

If I use my original equation I get.. T=m1gsin30-f / ma
T = (15kg)(9.80m/s^2)(sin30)-31.825N / (15kg) (a) <---and I don't know a?

16. Oct 6, 2012

### tiny-tim

block B ?

17. Oct 6, 2012

### SherBear

Could this finally be correct?
T = (15kg)(9.80m/s^2)(sin30)-31.825N / (15kg) (12kg)
T= .637 N ?

18. Oct 6, 2012

### tiny-tim

i'm confused

you had T - m1gsin30° - 31.825 = m1a

and m2g - T = m2a

to find T, you need to eliminate a, so multiply the first equation by m2 and the second equation by m1, and then subtract

19. Oct 6, 2012

### SherBear

Sorry Tim =/

T=m1g sin 30° - 31.825 (m2)

T=(15 kg) (9.80 m/s^2) (sin 30°) - 31.825 (12 kg)

T=1382.1 N?

Then...2nd Equation

T = m2g (m1)

T=(12 kg) (9.80 m/s^2) (15 kg)

T=1764 N?

Then Subtract the 2....

1764 - 1382.1 = 381.9 N

T= 381.9 N?

20. Oct 7, 2012

### SherBear

Here's a different way, can you tell me if it's right or wrong?

forces on A
u*m*g*Cos30 -T = ma
T - 0.25*15*9.8*cos30 = 15*a
T - 31.82N = 15a .........................................i

Forces on B
m*g-T = ma
12*9.8 -T = 12a
117.6-T = 12a...........................................ii

-31.82 + 117.6 = 27a
85.78 = 27a
a = 3.17 m/s/s
Now T = 117.6 - 12*9.8 = 0 No motion