# 2 blocks on a ramp with a pulley PLS.HELP

1. Jun 14, 2009

### brunettegurl

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

Two blocks,connected by a string over a pulley, slide along the frictionless ramp shown in the diagram. What is the tension in the string?

2. Relevant equations

$$\sum$$F=ma

3. The attempt at a solution
so i set up an equation for each mass
m1=3kg
$$\frac{Fg1+Ft}{m1}$$=a
m1=5kg
$$\frac{Fg2+Ft}{m2}$$=a

i equated the two since the a would be the same to solve for Ft
m2[(m1*g*cos30deg)+T1]=m1[(m2*g*cos60deg]+T1]

when i solve the equation and simplify for T i get 26.9N when i should be getting around 25 N..pls help

#### Attached Files:

• ###### blocks&string.jpg
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2. Jun 15, 2009

### tiny-tim

Hi brunettegurl!

(i can't see the diagram yet, but …)

if I'm guessing right, the accelerations aren't both a …

one is minus a

3. Jun 15, 2009

### Staff: Mentor

Also: Gravity acts down the incline while tension acts up--they must have different signs.

4. Jun 15, 2009

### brunettegurl

m1=3kg
$$\frac{-Fg1+Ft}{m1}$$=a
m1=5kg
$$\frac{Fg2-Ft}{m2}$$=a

i equated the two since the a would be the same to solve for Ft
m2[(-m1*g*cos30deg)+T1]=m1[(m2*g*cos60deg]-T1]
does that look better??

5. Jun 15, 2009

### tiny-tim

Hi brunettegurl!
Same thing applies to the other block.

6. Jun 15, 2009

### brunettegurl

so is my new equation wrong??

7. Jun 15, 2009

### tiny-tim

oops!

yes, the first two equations are correct …

but shouldn't the final one have sin rather than cos?

8. Jun 15, 2009

### brunettegurl

why would it be sin instead of cos??

Last edited: Jun 15, 2009
9. Jun 15, 2009

### ideasrule

Try splitting the force of gravity, a downward vector, into a component parallel to and a component perpendicular to the ramp. You'll see that the parallel component is mgsin(a), where a is the angle of inclination of the ramp.