# Two blocks on an Incline w/ Friction

1. Feb 3, 2010

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

Two blocks attached by a string (see figure) slide down a 20° incline. Block 1 has mass m1 = 0.65 kg and block 2 has mass m2 = 0.30 kg. In addition, the kinetic coefficients of friction between the blocks and the incline are 0.30 for block 1 and 0.20 for block 2.
(a) Find the magnitude of the acceleration of the blocks.
m/s2

(b) Find the tension in the string.
N

2. Relevant equations
Where do I go from here?

3. The attempt at a solution

Mass 1
Fg=9.81*.65=6.38 N
Fgx= Fgsin(20)
=6.38sin(20)=2.18 N
Fgy=Fgcos(20)
=6.38cos(20)= 5.995 N
Fkinetics=mk*FN
=.3*5.995=1.7985 N

Mass 2
Fg=9.81*.3=2.94 N
Fgx=2.94sin(20)=1.006 N
Fgy=2.94cos(20)= 2.76
Fkinetics= .2*2.76= .552 N

I believe I have all of the numbers needed, I just do not know how to solve for the answers using these numbers.

2. Feb 3, 2010

### PhanthomJay

Your numbers look good, so now first look at the 'system' of blocks and identify the magnitude and direction of all the external forces acting on the system in the direction parallel to the incline (you've already identified their magnitude). Then apply Newton's 2nd law to get the acceleartion. Then give a shot at finding the cord tension, using a free body diagram of one of the blocks to identify the forces acting on the individual block in the x direction.

3. Feb 3, 2010

Would it be:

1.006+2.18-.552-1.799=max?

I'm not sure what to use for the mass...would it be .65+.3?

4. Feb 3, 2010

### PhanthomJay

yes
Yes again. When you look at the system of the blocks, you must use the total mass of both. When you isolate just one of the blocks (as you must do for part b), then use the mass of that block alone to solve for the tension force in the cord connecting the 2 blocks.

5. Feb 3, 2010

I got:

ax=87.9 m/s2

But I have no idea what the equation is for tension...

6. Feb 3, 2010

### Eric_meyers

I'm guessing the second block is hanging over the triangle i.e. it's hanging vertically downward.

The way to solve for tension is to ask yourself two questions:

1.) If you removed the block hanging down vertically from the system - what would the tension be? What does this answer tell you about what's contributing to the tension in the rope?

2.) If the tension in the rope wasn't the same everywhere, i.e. if there was a force in the rope, that would imply the rope had acceleration wrt to itself, but this isn't the case: what does that imply about the tension force all along the rope?

7. Feb 4, 2010

### PhanthomJay

Your math is off by a factor of 100:
1.006+2.18-.552-1.799=m(a_x)
0.835 = 0.95(a_x)
a_x = 0.879 m/s^2
a_x = 0.88 m/s^2 (2 significant figures)
Isolate one of the blocks in a free body diagram; what forces act on that block in the x direction parallel to the incline? (the tension force will be one of them). Then use Newton 2 applied to that block alone, noting that it's acceleration is the same as the accleration of both blocks together, as long as T comes out to be a positive number. If T comes out negative, it's a whole new ballgame. Which block is which...you don't have a figure...which one is toward the bottom of the incline??