Does a massless string cause a normal force when attached to a wall?

Click For Summary
A massless ideal string does not create a normal force when attached to a wall, as it cannot exert a pushing force. The tension in the string is directed along its length and does not contribute to a normal force perpendicular to the wall. The normal component of force arises only when there is weight acting on the string, such as a hanging mass. If the mass is removed, the string becomes limp and exerts no force. Therefore, the presence of tension alone does not result in a normal force at the attachment point.
laser
Messages
104
Reaction score
17
Homework Statement
Does a massless string cause a normal force when attached to a wall?
Relevant Equations
F=ma probably
1702134312752.png

I'm almost certain that if it was a "steel rod" or something heavy like that, the normal (the force written in green) would exist. But does it exist for a "massless ideal string"? I mean, there is tension in the string of course, but would that cause the normal perpendicular to the wall?
 
Physics news on Phys.org
At the point of attachment the string exerts tension that is directed along the string and away from the point of attachment (you can't push with a string). This tension has a normal component and a component parallel to the surface. It's not the weight of the string that is the cause of this force but the weight of the ball hanging at the other end of the string that keeps the string taut. If you remove the ball, the string will go limp and exert no force if it is massless.
 
Would the cart shown in the attached diagram remain where represented when "there is tension in the string of course"?

Spehre cart.jpg
 
Thread 'Correct statement about size of wire to produce larger extension'
The answer is (B) but I don't really understand why. Based on formula of Young Modulus: $$x=\frac{FL}{AE}$$ The second wire made of the same material so it means they have same Young Modulus. Larger extension means larger value of ##x## so to get larger value of ##x## we can increase ##F## and ##L## and decrease ##A## I am not sure whether there is change in ##F## for first and second wire so I will just assume ##F## does not change. It leaves (B) and (C) as possible options so why is (C)...

Similar threads

  • · Replies 10 ·
Replies
10
Views
5K
Replies
17
Views
3K
  • · Replies 3 ·
Replies
3
Views
984
Replies
8
Views
953
  • · Replies 12 ·
Replies
12
Views
5K
  • · Replies 9 ·
Replies
9
Views
3K
  • · Replies 3 ·
Replies
3
Views
6K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 4 ·
Replies
4
Views
3K
Replies
3
Views
3K