Solving for Tension in Rope: A & B Blocks, Theta=30°, us=.25

[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!

 

[tiny-tim]

Hi SherBear!

(try using the X₂ button just above the Reply box )

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?

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

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

Block B
F=m2g-T2=m2a

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 ​

 

[SherBear]

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

 

[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)

 

[SherBear]

Yes, I did get email notifications both times for your answer :)

 

[SherBear]

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

 

[SherBear]

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

 

[tiny-tim]

(hmm … i got both of those notifications! )

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

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)

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

depends which way you're measuring a …

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

 

[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?

 

[tiny-tim]

looks ok

 

[SherBear]

looks ok

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

 

[tiny-tim]

uhh?

f is the friction​

 

[SherBear]

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

 

[tiny-tim]

use your original equation …

t1-m1gsin30°-f=ma

 

[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?

 

[tiny-tim]

block B ?

 

[SherBear]

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

 

[tiny-tim]

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

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

 

[SherBear]

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

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?

 

[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

adding i and 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

 

[SherBear]

So any input? I still don't know if I'm doing this right.

 

[tiny-tim]

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

adding i and 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

yes, that's the right method, but in A you've missed out the m₁gsin30°

(alternatively, if you only want T and not a, then you could eliminate a instead of T )

 

[SherBear]

I don't like that way, could I also find Acceleration and knowing both masses I could rewrite:

the sum of F=m2g-T=m2a as
T= m2g / m2a = 12(9.8) / 12 (.454 assuming the acceleration i computed is correct) = 21.58 N

T=21.58 N ?

 

[SherBear]

Or for box b use sum of F=T m2g so T=m2g...12(9.8 m/s^2) = 117.6N ?

 

[SherBear]

Which T is the same all the way through because it's 1 rope?