Four Firefighters Holding a Net: Calculating Net Force Question

[basenne]

Homework Statement

Four firefighters hold a square net, one at each corner. Each person exerts a force of 190 N whose line-of-action passes through a point just below the center of the net and makes an angle with the vertical of 50°. What is the net force the firefighters exert on the net?

Homework Equations

F=ma

The Attempt at a Solution

I've tried splitting the problem into four vectors and solving from there, but I cannot seem to get the correct answer. Also, don't the forces cancel out?

I guess what I'm trying to say is that I'm lost. Any indication on how to start this one would really be appreciated.

 

[Heimisson]

Always in these force problem start by drawing a picture. Take every force and calculate the magnitude in the x-axis direction and y-axis direction by taking the cos and sin of the angle with x-axis. Sum the x direction forces together and be sure to take in account that if they have opposite direction they can cancel. When you have the total force in x-direction and y-direction you get the net force by using the Pythagoras theorem that is if the forces in either x or y don't cancel out.
I have a feeling the answer is 4*190*sin(40).

But understand that x and y directions a chosen by you and are completely relative.

 

[basenne]

Yes, your answer is correct. However, I'm still confused with how you arrived at that conclusion. I solved the vectors (perhaps incorrectly?) and due to direction, all of the forces seem to cancel.

 

[Heimisson]

However, I'm still confused with how you arrived at that conclusion.

By doing a lot of these problems.


OK be careful they say the angle is 50 with vertical. Usually we prefer the angle with horizontal (x-direction) that is 40.

So you have 4 fireman opposite each other. They are all using the same force but different directions. The net is a square but we shall think of it as a clock(because I can't draw a picture here): so the fireman a positioned at 12, 3, 6 and 9

The one at 12 pulls cos(40)*190N in horizontal (x) but so does the one in 6 but in opposite direction so the forces cancel. The same goes for the ones at 9 and 3. However there is no fireman under the net pulling in the vertical direction. So the vertical part of the force (for every fireman) sin(40)*190N does not cancel. Meaning the total force is the sum of these.


I hope this helps.

 

[basenne]

How, exactly, do we know that there is no fireman under the net pulling in the vertical direction? Isn't the vertical pull sin(40)*190N also?

 

[Heimisson]

How, exactly, do we know that there is no fireman under the net pulling in the vertical direction? Isn't the vertical pull sin(40)*190N also?

Yes the vertical pull is sin(40)*190N the horizontal is cos(40)*190N.

Because they say: Four firefighters hold a square net, one at each corner.

I don't think i can explain this any better in written text. Just try to imagine, draw a picture, make a force diagram or something of what's going on.

 

[basenne]

Oh! I get it! Thanks a lot!