Forces (no friction) help please

In summary, the tension in the cable below the worker is 1594.9 N and the tension in the cable above the worker is 2573.74 N. The tension in the cable above the worker is calculated by adding the weight of the worker and the crate and multiplying it by the sum of the acceleration and the force of gravity.
  • #1
MakeItThrough
13
0

Homework Statement



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?

Homework Equations


Force = mass x acceleration
weight = mass x g

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.
 
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  • #2
MakeItThrough said:

Homework Statement



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?

Homework Equations


Force = mass x acceleration
weight = mass x g

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.

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
MakeItThrough said:

Homework Statement



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?

Homework Equations


Force = mass x acceleration
weight = mass x g

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.

Pretend the crate and the worker are stationary. What would you say the tension on the cable above the worker would be?
 
  • #4
Ignea_unda said:
Pretend the crate and the worker are stationary. What would you say the tension on the cable above the worker would be?

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
MakeItThrough said:
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?

Correct.

Now if we want to accelerate that worker, what needs to happen? (Remember the box is attached to him)
 
  • #6
Ignea_unda said:
Correct.

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

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
MakeItThrough said:
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?

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
cepheid said:
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.

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

1. What is a force?

A force is a push or pull that can cause an object to accelerate or change its direction of motion. It is a vector quantity, meaning it has both magnitude and direction.

2. What are the different types of forces?

There are four fundamental forces in nature: gravity, electromagnetism, strong nuclear force, and weak nuclear force. Other types of forces include tension, compression, friction, and applied force.

3. How are forces related to motion?

According to Newton's first law of motion, an object at rest will remain at rest and an object in motion will continue to move at a constant velocity unless acted upon by an external force. Forces can change the speed, direction, or shape of an object's motion.

4. Can forces cancel each other out?

Yes, forces can cancel each other out if they are equal in magnitude and opposite in direction. This is known as equilibrium and is observed when the net force on an object is zero.

5. How do forces affect the motion of an object on an inclined plane?

On an inclined plane, the force of gravity is split into two components: the normal force perpendicular to the surface and the force of gravity parallel to the surface. The object will accelerate down the plane due to the force of gravity and friction may also play a role in its motion.

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