Tension and acceleration in pulley system with inclined and horizontal blocks

[ViewtifulBeau]

Block1 has a mass of 8.63 kg and block2 has a mass of 1.24 kg. b1 is on an inclined plane with no friction. The plane is 30 degrees from the horizontal. b1 is connected to b2 with a string-pully system (no mass or friction). b2 is on a horizontal plane witha kinetic friction of .145 Block m1 is sliding down, pulling block m2. I have to find the tension in the string and the acceleration of the blocks. so i did this:

(mass(b1) * 9.81 * sin 30) + (mass(b2) * 9.81 * .145) = T

to find the tension which equals 44.09 which is not the answer. I can find the acceleration once i get the tension i believe by f=ma.

 

[Doc Al]

The most straightforward way to solve this problem is by separately applying Newton's 2nd law to each block. Combine those equations to solve for the tension and acceleration, the two unknowns. (Since the blocks are connected, they both have the same acceleration.)

 

[ViewtifulBeau]

i tried F=ma and i got 22.045 N and it still doesn't work

 

[Doc Al]

Did you do what I suggested? Apply F=ma to each block separately. You'll get two equations.

 

[ViewtifulBeau]

yeah i did it and i got :

(U*g*m(2) + m(1)*g* sin(theta))/2 = T

so it was (1.7638 + 42.37) /2 = 22.04 = T

but it is not right

 

[Doc Al]

Analyze each block separately:
(a) Identify the forces on m1; Apply F = ma to m1.
(b) Identify the forces on m2; Apply F = ma to m2.

You'll get two equations. Do this.