How do I find the horizontal force/traction force of this?

In summary, the conversation discusses a setup commonly used in hospitals to apply traction to an injured leg. The force of tension in the rope supporting the leg is found to be 78.5 N, and the conversation then moves on to determining the traction force exerted on the leg. It is explained that for a frictionless and ideal pulley system, the tension in the rope on one side is the same as the other side. The traction force is defined as the total horizontal component of force on the foot, and the conversation concludes with the question of how to find this force, with the expected answer being 105 N.
  • #1
riseofphoenix
295
2
8. A setup similar to the one shown in the figure below is often used in hospitals to support and apply a traction force to an injured leg.

4-p-031.gif


(a) Determine the force of tension in the rope supporting the leg.

F = mg
F = (8.00)(9.81)
F = 78.48
F = 78.5 N

Is that all I had to do? It says I got the answer for this part right.

(b) What is the traction force exerted on the leg? Assume the traction force is horizontal.

This is where I'm completely lost and confused.
First of all, the horizontal traction force they are referring to is the longest one on top right?
If so, then do I have to find the value of that horizontal line?
How? o_O
They say the answer is supposed to be 105 N

Help!
 
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  • #2
riseofphoenix said:
8. A setup similar to the one shown in the figure below is often used in hospitals to support and apply a traction force to an injured leg.

4-p-031.gif


(a) Determine the force of tension in the rope supporting the leg.

F = mg
F = (8.00)(9.81)
F = 78.48
F = 78.5 N

Is that all I had to do? It says I got the answer for this part right.
'Looks right to me! :approve:

If you still have questions why this is, you should assume that the system is in static equilibrium (i.e. nothing is moving). Second, assume that the pulleys are frictionless.

For a frictionless, simple pulley (with just one rope hanging off of it) in static equilibrium, the tension in the rope on one side is the same as the other side. [Edit, and assume the rope is ideal, mass-less, too.]
(b) What is the traction force exerted on the leg? Assume the traction force is horizontal.

This is where I'm completely lost and confused.
First of all, the horizontal traction force they are referring to is the longest one on top right?
If so, then do I have to find the value of that horizontal line?
How? o_O
They say the answer is supposed to be 105 N

Help!
What is meant by "traction force" is the total horizontal component of force on the foot, from the pulley attached to the foot. Notice that the pulley attached to the foot is being "pulled" to the right a little by the rope. (Well, technically it is still in static equilibrium due to the friction of the patient on the attached bed. But that's not the important thing. The point is that there is still a horizontal component of force acting on the foot-pulley ultimately from the rope.) This force acts to pull the foot away from the rest of the leg, in a net horizontal direction. Find this horizontal component of that force.
 
Last edited:

Related to How do I find the horizontal force/traction force of this?

1. How do I find the horizontal force/traction force of this object?

To find the horizontal force or traction force of an object, you will need to know the mass of the object and the coefficient of friction between the object and the surface it is on. Once you have this information, you can use the formula F = μN, where F is the horizontal force, μ is the coefficient of friction, and N is the normal force.

2. What is the normal force and how does it affect the horizontal force/traction force?

The normal force is the force exerted by a surface on an object that is in contact with it. This force is perpendicular to the surface and is equal in magnitude to the force exerted by the object on the surface. The normal force affects the horizontal force/traction force by determining the maximum amount of friction that can be exerted on the object.

3. Can I use the same formula to find the horizontal force/traction force in different situations?

Yes, the formula F = μN can be used to find the horizontal force/traction force in any situation as long as you know the mass of the object and the coefficient of friction between the object and the surface it is on. However, keep in mind that the coefficient of friction may vary depending on the surface and the object in question.

4. How does the angle of the surface affect the horizontal force/traction force?

The angle of the surface can affect the horizontal force/traction force by changing the normal force. If the object is on a flat surface, the normal force will be equal to the weight of the object. However, if the surface is angled, the normal force will be less than the weight of the object, resulting in a lower maximum friction force.

5. Are there any other factors that can affect the calculation of horizontal force/traction force?

Yes, there are other factors that can affect the calculation of horizontal force/traction force. These include the type of surface, the roughness of the surface, and the velocity of the object. These factors can impact the coefficient of friction and therefore, the resulting horizontal force/traction force.

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