Physics problems?/?

Question

two masses, m1 and m2, connected by a massless string, are accelerated uniformly on a frictionless surface. What is the ration of the tensions T1/T2.
I get an answer of m1/(m1+m2), but i am not sure if this is correct. Can you show me why this is correct?

Two masses, m1=12 kg and m2=35kg, are held connected by a massless rope hung over a frictionless light pulley. If the masses are released, what is the magnitude of the tension in the string.
My instructor says that the answer is 175N, but i don't get this answer. I get 225N. Why is this wrong. What i am doing is subtracting 12 from 35 and multpilying by 9.8. What is wrong with this?

Three blocks are connected by massless cords and rest on a frictionless horizontal surface. The blocks are pulled to the right. Mass m1=2, m2=3m3, with m1 on the left, and m3 on the right. IF the pulling force is equal to 90N, what is the tension in the cord between m1 and m2.

Why is the answer 49N?

Answer 1

I assume T2 is the string pulling both masses.
m1---(T1)----m2----(T2)---->F
From F=ma,
T2=F(total)=m(total)a.
m(total)=m1+m2
T2=(m1+m2)a
acceleration is constant for all equations involved due to the string.
T1=m1a
T1 is unaffected by m2, as its only constraint is an acceleration of a which will result in the force.
By simply dividing T1/T2
T1/T2=(m1a)/[(m1+m2)a]
the a's cancel to give T1/T2=m1/[m1+m2]


part 2
O
/ \
/ \
| |
| |
m1 m2

F=ma
m=-9.8
F1=12*-9.8=-117.6
F2=35*-9.8=-334
T=|F1-F2|=225
However if you were doing this quickly, and punched 30 in your calculator instead of 35, you end up with 176.4N. I would guess that's what happened.

Last part
Are you sure m2=3m3? No relations given between m1 and the other two? Here are some calculations, but no results.
m1---(T)---m2----m3----->90N

F=ma

F(total)=90N=m(total)a

F1=T=m1a
or by rearranging
a=T/m1=T/2

m(total)=m1+m2+m3= 2 +3m3 +m3 = 2+4m3
plug into first equation

F(total)=90=m(total)a=[2+4m3][T/2]

90=[2+4m3][T/2]
180=[2+4m3]T

Now from
90=[2+4m3]a
a = 90/[2+4m3]
and from earlier
a=T/2

we have
T/2=90/[2+4m3]
T=180/[2+4m3]
same result as earlier
I don't think a solution is possible without further information about m3.

Answer 2

The first part is correct, do a force balance on each mass,
yo get for the first mas M1a=t1, for the second mass you get m2a=t2-t1, pluggin in a=t1/m1 and rearranging gives yout t1/t2.

For the second part, very similar, the key is setting up a free body diagram for eachfor the first mass you get M1a=T-M1g; for the second mass you get -M2a=T-M2g, solve for T.

the third problem is very similar to the first two, I think if you can do the first two you should get the third pretty easily.

remember the key is the free body diagram for each mass.