Hooke's Law: Combining Forces & Question on Resultant Force

[Zynoakib]

Homework Statement

Hooke’s law describes a certain light spring of
unstretched length 35.0 cm. When one end is attached
to the top of a doorframe and a 7.50-kg object is hung
from the other end, the length of the spring is 41.5 cm.
(a) Find its spring constant. (b) The load and the spring
are taken down. Two people pull in opposite directions
on the ends of the spring, each with a force of 190 N.
Find the length of the spring in this situation.

I am having a small problem on (b)

Homework Equations

No

The Attempt at a Solution

Since the question says "Two people pull in opposite directions on the ends of the spring, each with a force of 190 N.", shouldn't the resultant force be 190N x 2 instead of just 190N (the answer)?

 

[BvU]

Look at the end points of the spring separately and make a drawing of the forces. Newton (action = -reaction) applies. In a as well as in b. (what do you think is the force on the doorframe in a ?)

 

[Chestermiller]

The resultant force is the magnitude of the
two forces. The relevant equation in a situation like this is the sum of the forces.

Fz=(Fx^2+Fy^2)^1/2

We can neglect Fy on this occasion, treating Fx as a vector its magnitude is,

Fx= (Fa^2+(-Fb)^2)^1/2 N
= (190^2 + (-190)^2)^1/2 N
= 268.7005769 N

The correct notation for the equation is sigma Fx, Fy, Fz. Now we know how much force is being exerted on the spring we can use hooks law,

F=-kX
X=F/-k

I'd have a look into statics with free body diagrams there great for visualising problems.

Since you're so good with free body diagrams, please show us your free body diagram for this situation. I'm guessing that your interpretation of the problem statement is very different from that of the rest of us.

Chet

 

[Zynoakib]

Look at the end points of the spring separately and make a drawing of the forces. Newton (action = -reaction) applies. In a as well as in b. (what do you think is the force on the doorframe in a ?)

The force on the doorframe in a should be 7.5 x 9.8 = 73.5 N

That's what I can imagine.

 

[BvU]

The force on the doorframe in a should be 7.5 x 9.8 = 73.5 N

That's right (plus some force from the weight of the spring itself).

(By the way, learn yourself to always include units -- and check them -- like: 7.5 kg $\times$ 9.8 m/s² = 73.5 N -- and N = kgm/s² )​

So for the spring constant, I hope you divide this 73.5 N by the extension (0.065 m) and not by twice 73.5 N.

Can you see the similarity with the two people pulling in opposite directions ?

I agree with your picture. Wholeheartedly: no matter what happens on the other end of the spring. It may be attached to the side of a doorframe (which then exerts a force of 190 N on the spring ! -- a mirror picture of yours) or it may be that someone holds the other end in place. If there's a curtain hanging in the middle, I won't be able to tell the difference !

Same with a rope: if the tension in the rope is 190 N, then a force of 190 N is required on both ends. Opposite and equal, so that the vector sum is zero: no acceleration (another Newton law). Again: doorframe or person -- doesn't matter.

 

[Zynoakib]

Thank you, I get it now!