Calculating the force at a single point of contact

In summary, the conversation discusses the feasibility of a chrome wire with a diameter of 0.5 mm and a tensile strength of 600 N/mm^2 to withstand a force of 2 N while holding a nylon wire under tension. The calculations show that the chrome wire should easily hold the fishing line and even applying a force at almost right angles is more likely to break the fishing line than the wire. It is also mentioned that the wire may cut the fishing line. The possibility of a tensor matrix being involved is also brought up.
  • #1
hamzaaaa
31
0
Hi all

May be someone can help me with this problem.

There is a isa-chrome60 wire (d=0.5 mm, tensile strength=600 N/mm^2) which is held at both ends using end connectors ( length of the wire = 20mm). And there is a nylon wire (d=0.5 mm) which is under tension (because of torsion spring) and it is held by chrome wire at a single point. Say the force acting on the fishing line is 2 N. I would like to know if it would be possible to know that the chrome wire could withstand this much of force or not?
Is there any mathematical way to do that?

Thanks in advance.
 
Engineering news on Phys.org
  • #2
There is a isa-chrome60 wire (d=0.5 mm, tensile strength=600 N/mm^2) which is held at both ends using end connectors ( length of the wire = 20mm). And there is a nylon wire (d=0.5 mm) which is under tension (because of torsion spring) and it is held by chrome wire at a single point.

Just on those figures,
the cross sectional area of the chrome wire is pi * .25 * .25 = 0.196 sq mm
So the tensile strength should be 600 * 0.196 or 118 N
This is about the weight of 11 Kg.

2 N is about the weight of 200 grams.

So, yes the wire should easily hold the fishing line in a straight line and even applying a force at almost right angles is more likely to break the fishing line than the wire.

As we discussed elsewhere, the wire may cut the fishing line, though.
 
  • #3
Thanks again vk6kro..
I had done similar calculations, but wanted to confirm if it was the right way. I thought some tensor matrix might get involved.

Thanks and Regards
 

Related to Calculating the force at a single point of contact

1. What is the formula for calculating force at a single point of contact?

The formula for calculating force at a single point of contact is F = m x a, where F represents force, m represents mass, and a represents acceleration.

2. How do you determine the direction of the force at a single point of contact?

The direction of the force at a single point of contact can be determined by using a coordinate system and assigning positive and negative values to the x and y axes. The direction of the force will depend on the signs of the x and y components of the force.

3. Can you calculate the force at a single point of contact for an object in motion?

Yes, the force at a single point of contact can be calculated for an object in motion by using the formula F = m x a, where m represents the mass of the object and a represents the acceleration of the object.

4. How do you account for friction when calculating the force at a single point of contact?

Friction can be accounted for by using the coefficient of friction in the formula for force at a single point of contact. The formula becomes F = u x N, where u represents the coefficient of friction and N represents the normal force.

5. Are there any other factors that need to be considered when calculating force at a single point of contact?

Yes, other factors such as air resistance, surface conditions, and the shape of the object can also affect the force at a single point of contact and should be taken into account in the calculation.

Similar threads

Replies
4
Views
712
  • Mechanical Engineering
Replies
26
Views
2K
  • General Engineering
Replies
15
Views
2K
  • Introductory Physics Homework Help
Replies
18
Views
2K
  • Engineering and Comp Sci Homework Help
Replies
1
Views
4K
  • Electrical Engineering
Replies
1
Views
2K
  • Programming and Computer Science
Replies
26
Views
3K
  • Mechanical Engineering
Replies
1
Views
2K
  • Engineering and Comp Sci Homework Help
Replies
9
Views
2K
  • Special and General Relativity
3
Replies
75
Views
3K
Back
Top