Acceleration on a pulley system with 3 blocks

[mackeis]

Homework Statement

Three objects are connected on an inclined table. The objects have masses 8.0 kg, 4.0 kg and 2.0 kg as shown, and the pulleys are frictionless. The tabletop is rough, with a coefficient of kinetic friction of 0.47, and makes an angle of 20.0° with the horizontal. The 8.0kg block hangs on the left, the 4.0kg block in the middle on the table (20.0° t the left and above horizontal), and the 2.0kg is hanging on the right.

Draw a free-body diagram for each object. (did this but can't display on the forum)
Determine the acceleration and direction of motion of the system.
Determine the tensions in the two cords.

Homework Equations

εF=ma
trig
T-mg=ma

The Attempt at a Solution

I can do this with two blocks but 3 on a angle through me off.
mg for the 8.0kg block =78.4N and mg for the 2 kg block is 19.6kg

I know that for the block on the table:
Fgx = sin 20 (39.2) = 13.4N
Fgy = cos 20 (39.2)= 36.83N - Force normal and Fgy are equal
Ff=0.47(36.83)

The using εFx=max
19.6=13.4-17.3-78.4 = (14)a
a= -4.5m/s/s or 4.5 m/s/s moving to the left.

Not sure if I did this right and when calculating tension in a 3 block pulley system on a angle.

Thanks

 

[Doc Al]

The using εFx=max
19.6=13.4-17.3-78.4 = (14)a

It's a bit unclear what you are doing here.

You need equations for all three masses, which you will solve together to get the acceleration.

 

[mackeis]

Sorry, first time using this forum. What i did was
εFx=max and i used mg for the masses on either end of the system
so in this case i had
εFx = m3g+Fgx-Ff-m2g=(m1+m2+m3)a
εFx = 8x9.8 + sin20(4x9.8) - .47xcos20(4x9.8) - 2x9.8 = (8+4+2)a
19.6+13.4-17.3-78.4 = (14)a

The Fgx and Ff apply to the box on a slant on the table.

 

[Doc Al]

εFx=max and i used mg for the masses on either end of the system

I think what you did was assume that the tension in each rope equals the weight of the hanging mass. You can't do that!

If the tension equaled the weight of the masses, then the net force on each hanging mass would equal zero. No acceleration!

Instead, call the unknown tensions T1 and T2 (or whatever). You need to solve for the three unknowns: the acceleration and the two tensions. That's why you need three equations, one for each mass.

 

[HalfLostAndSearching]

for providing your work and thought process! Your free-body diagrams and use of Newton's second law (ΣF=ma) are correct. However, there are a few things to consider when solving this problem:

1. The acceleration of the system will be the same for all three blocks, since they are connected by a single string. Therefore, your calculation for acceleration should use the total mass of the system (8.0kg + 4.0kg + 2.0kg = 14.0kg).

2. When calculating the net force in the x-direction, you must also consider the tension in the string connecting the 8.0kg block and the 4.0kg block. This tension will be equal to the tension in the string connecting the 4.0kg block and the 2.0kg block.

3. To find the tensions in the strings, you can use the fact that the net force in the y-direction for each block must be zero (since they are not accelerating in the y-direction). This means that the tension in the string connected to the 8.0kg block must be equal to the weight of the block (78.4N), and the tension in the string connected to the 2.0kg block must be equal to the weight of the block (19.6N).

Using these considerations, you should be able to correctly solve for the acceleration and tensions in the system. Keep up the good work!