# Forces (no friction) help please

1. Sep 27, 2011

### MakeItThrough

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

http://www.webassign.net/CJ/04_75.gif
A cable is lifting a construction worker and a crate, as the drawing shows. The weights of the worker and crate are 922 and 1500 N, respectively. The acceleration of the cable is 0.620 m/s2, upward.
(a) What is the tension in the cable below the worker?
1594.9 N
(b) What is the tension in the cable above the worker?

2. Relevant equations
Force = mass x acceleration
weight = mass x g
3. The attempt at a solution
(a) What is the tension in the cable below the worker?
1594.9 N
(b) What is the tension in the cable above the worker?

I have figured out part a, here is my work:
Since it is accelerating upward, force of tension is greater than the weight.
1500 N = mass x 9.8 m/s^2
mass = 153.1 kg
acceleration = 0.620 m/s^2
FT - 1500 = 153.1(0.620)
FT = 1594.9 N
I used this same method for B, but the answer is wrong. How do I get the tension of the cable above the worker? Thanks a bunch in advance.

2. Sep 27, 2011

### cepheid

Staff Emeritus
Draw a free body diagram for the worker. You'll find that, unlike the crate, the worker has three forces acting on him. One of those is obviously his weight. As for the other two, well, bear in mind that the work has a cable above him and a cable below him, and both of them are in tension.

3. Sep 27, 2011

### Ignea_unda

Pretend the crate and the worker are stationary. What would you say the tension on the cable above the worker would be?

4. Sep 27, 2011

### MakeItThrough

If they are stationary, then acceleration would be zero and the net force would also be zero. So the tension of the cable above the worker would be equal to all the forces that are pushing the worker down. Right?

5. Sep 27, 2011

### Ignea_unda

Correct.

Now if we want to accelerate that worker, what needs to happen? (Remember the box is attached to him)

6. Sep 28, 2011

### MakeItThrough

The acceleration is 0.620 m/s^2 upward. The tension of the cable above the worker the must be equal to the worker's mass AND also the mass of attached box below him times the acceleration.
Ft = 94 + 153 x (0.620)
is this correct?

7. Sep 28, 2011

### MakeItThrough

that must be wrong then. The result I get is 153.14. The difference seems so little compared to my answer for part A...? What am i missing????????

8. Sep 28, 2011

### MakeItThrough

Of course! Gravity!! How could I forget? I'm not sure whether to add or subtract 0.620 from 9.8. Gravity is pulling down, while the acceleration of 0.620 is going upward. So it makes sense to subtract.
FT = (94 kg + 153 kg) x (9.8 - 0.620) = 2267.46
???
*Edit
That answer was wrong. So now I have to add 0.620 to 9.8. And the correct answer is
Force = Mass x Acceleration
FT = (94 kg + 153 kg) x (9.8 m/s/s + 0.620 m/s/s) = 2573.74 N
That was the right answer for part B. What I don't understand is why do I have to add the acceleration and not subtract it?

9. Sep 28, 2011

### cepheid

Staff Emeritus
This is why it is best to draw a free body diagram and set the sum of forces to be equal to the net force. It automatically takes into account the signs of everything properly.

In this case:

Ftension + Fgrav = Fnet

Using the convention that upwards is positive and downwards is negative, this becomes:

Ftension - mg = ma

or:

Ftension = ma + mg = m(a + g)

That's why you add the accelerations. It just comes from applying Newton's second law (sum of all forces = ma) the way I just did above.

This result makes sense. The tension force has to be larger than the combined weight of the two loads, because there is an upward acceleration and hence there must be a net upward force. The net upward force is because the tension force is pulling upwards more than the gravitational force pulls downwards.

10. Sep 28, 2011

### MakeItThrough

thanks, i'll remember from now on to draw a free body diagram for Forces