Pulley Mechanics: Equal Forces on Spring Scale?

In summary, the weights on each setup exert equal forces on the spring scale, neglecting the weight and friction of the pulleys.
  • #1
keemosabi
109
0

Homework Statement


A 100 gram mass is hanging vertically from a spring scale and exerts a certain force. Another 100 gram mass is set up in the following situation (there is a pulley, spring scale, mass, and ring stand holding up the spring scale):
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Do the weights on each setup exert equal forces on the spring scale, neglecting the weight and friction of the pulleys?



Homework Equations





The Attempt at a Solution


I thought that I would have to separate the second one into vectors to figure out if there is a difference.
 
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  • #2
keemosabi said:

Homework Statement


A 100 gram mass is hanging vertically from a spring scale and exerts a certain force. Another 100 gram mass is set up in the following situation (there is a pulley, spring scale, mass, and ring stand holding up the spring scale):
Do the weights on each setup exert equal forces on the spring scale, neglecting the weight and friction of the pulleys?

The Attempt at a Solution


I thought that I would have to separate the second one into vectors to figure out if there is a difference.

What is the tension in the string that holds the weight in each situation?

In the one the tension goes straight up to the scale.
In the other if there is a difference in the tension at the weight and the tension at the scale, what will the weight be doing? Will it be in motion?
 
  • #3
When we did the experiment the only measurement we were told to make was reading the spring scale. In the first situation the spring scale read .98 N, while in the second situation the scale read .8N. However, my teacher said that the readings should have been the same the same in both situations, but that doesn't make sense.
 
  • #4
keemosabi said:
When we did the experiment the only measurement we were told to make was reading the spring scale. In the first situation the spring scale read .98 N, while in the second situation the scale read .8N. However, my teacher said that the readings should have been the same the same in both situations, but that doesn't make sense.

That's where friction and the real world intrudes on things.

Oil the pulley next time?
 
  • #5
Thank you for the reply.

So the measurement in Newtons should have been the same both times? Why is that? Don't you need to break down the force that the 100 g mass exerts into vectors to get the x any y directions of force?
 
  • #6
keemosabi said:
Thank you for the reply.

So the measurement in Newtons should have been the same both times? Why is that? Don't you need to break down the force that the 100 g mass exerts into vectors to get the x any y directions of force?

The only direction of real interest with tension is the direction of the cable.
 
  • #7
Wouldnt the mass being hung at an angle from the spring scale decrease the amount of force it exerts on the scale in horizontal direction?
 
  • #8
keemosabi said:
Wouldnt the mass being hung at an angle from the spring scale decrease the amount of force it exerts on the scale in horizontal direction?

The weight is hanging straight down.

What angle?

The force of the weight is along the tension translated back to the scale.

If the scale is supporting with less force than the weight ... is the cable growing, is the weight in motion? Otherwise the sum of the forces are 0.
 
  • #9
LowlyPion said:
The weight is hanging straight down.

What angle?
The angle between the weight and the spring scale. The hypotenuse of the string in the vertical and horizontal directions.

The force of the weight is along the tension translated back to the scale.
So the force of the weight is translated up through the cable, and then horizontally to the scale? Making the force that the weight exerts the sum of the tension of the strings in the vertical and horizontal directions?

If the scale is supporting with less force than the weight ... is the cable growing, is the weight in motion? Otherwise the sum of the forces are 0.
The system is at equilibrium.

Thank you for your help thus far.
 
  • #10
keemosabi said:
So the force of the weight is translated up through the cable, and then horizontally to the scale? Making the force that the weight exerts the sum of the tension of the strings in the vertical and horizontal directions?

Sorry it doesn't work that way.
 
  • #11
LowlyPion said:
Sorry it doesn't work that way.
So then why do you not need to break down the weight's pull into vectors?
 

1. How do pulleys change the amount of force needed to lift an object?

Pulleys allow for a change in the direction of the applied force, which can make it easier to lift an object. By using multiple pulleys and changing the direction of the force, the overall force needed to lift an object can be reduced.

2. What is the mechanical advantage of using pulleys?

The mechanical advantage of pulleys is the ratio of the output force to the input force. By using pulleys, the mechanical advantage can be increased, making it easier to lift heavy objects.

3. How does the number of pulleys affect the force required to lift an object?

The more pulleys that are used, the less force is required to lift an object. This is because each additional pulley adds to the mechanical advantage, reducing the overall force needed.

4. Can pulleys reduce the force needed to move an object horizontally?

Yes, pulleys can also be used to move an object horizontally. By using multiple pulleys and changing the direction of the force, the overall force needed to move the object can be reduced.

5. Is there a limit to how many pulleys can be used to lift an object?

Technically, there is no limit to the number of pulleys that can be used to lift an object. However, as more pulleys are added, the efficiency of the system decreases due to increased friction and the weight of the pulleys themselves. Therefore, there is a practical limit to the number of pulleys that can be used effectively.

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