Balancing Forces: Solving a Physics Problem with Trigonometry

[freerider78]

Hi, I hope this type of question is against the rules here. But I have a slight emergency. I have to substitute teach a HS physics class tomorrow. I am a biology teacher, so this is out of my realm. I have to explain a couple problems and have no idea how to begin. I am figuring I need to sin and cos some angles, but the rest is unknown. Here's the question, if anyone can step me through the math of this, I will be forever grateful
A force board is set up in a physics lab and the three spring scales are hooked to the central ring and the notched perimeter of the table. The first scale reads 12.5 N at an angle of 0 degrees. The second scale reads 17.3 N at an angle of -34 degrees. If the three forces are balanced, what are the reading and angle of the third spring scale?

 

[Ja4Coltrane]

This really isn't clear--could you describe a picture?
I mean, what is the "first scale" where are the others? what does -34 degrees mean?

 

[freerider78]

Thanks for replying, but I'm sorry, that is all the info I have. I assume I would draw an x and y axis, with the 0 degree vector at the horizontal x and the 34 degree one below the x-axis. Like if it was a clock, it would be 3:20. Does this help?

 

[Ja4Coltrane]

okay, if they are balanced forces, then the sum of all forces is equal to zero. This means that if you add the force from the spring pulling horizontally to the force of the -34 degree spring IN THE X DIRECTION you should find the x- component of the other spring.
so...let me find my calculator...
here is the idea--the sum of all forces in the x direction is equal to zero and the same thing holds ture for y. So if you notice, if the angle of -34 degrees was changed to -15 degrees, you would expect the force in the x direction to be greater. you notice that as cos@ moves from -34 to -15 it also becomes greater, so you could use this logic to determine that the x-component of the force excerted by that spring is (17.3N)cos34.

 

[Ja4Coltrane]

now add that value to the 12.5 and you know how hard the other spring is pushing to the left.

 

[Ja4Coltrane]

now for the y direction you do the same thing, but you can completely ignore one of the springs. (17.5N)sin34=the force by the third spring upward.

 

[Ja4Coltrane]

the third spring pushes with two different forces in two perpendicular directions (up and to the left) so you can use a^2+b^2=c^2 to figure out the resultant force.

 

[Ja4Coltrane]

now use trig to find the angle--just draw it--tan@=opp/adj=Fy/Fx. So use tan^-1(Fy/Fx) to get the angle above the negative x-axis which should be subtracted from 180 to stay consistent. I hope that clears it up.

 

[Astronuc]

Ah! This is a typical net force = zero problem in x-y (Cartesian coordinates), i.e. statics.

Sum of forces in x-direction = 0, and
Sum of forces in y-direction = 0

For the first spring (12.5 N), it is entirely in the x-direction, and for the second spring (17.3 N), the component in the x-direction is given by 17.3 N cos (-34°).

The x-component of the third spring must be equal and opposite the sum of these two.

For the y-direction, the first spring has no component, while the second spring has component 17.3 N sin (-34°). The third spring must have a component of equal magnitude, but in the opposite (+y) direction.

See - http://hyperphysics.phy-astr.gsu.edu/hbase/vect.html#veccon

These may be of use in the future -
http://hyperphysics.phy-astr.gsu.edu/hbase/exprob/exmec.html

 

[freerider78]

Ja4Coltrane, thanks so much. I'm going to run through this and see if I comprehend it fully. It's still a little above my head, but the answer is right according to my calculations versus the answer sheet (give or take a few decimal places).

 

[Ja4Coltrane]

good!
yeah the trick is to see that every force in any direction can be thought of as two diferent forces in x and y directions.