Math question? system of equations involving trig

[hsbeth]

Daughter is stuck on a Physics problem.

 

She's got a system of equations with 2 unknowns:

 

Tcos(x) = 980

 

[Tsin(x)] = [6.67 * 10 ^ (-7)] / [1- 400tan(x) + 40,000tan^2(x) ]

 

x is in degrees, not radians

 

She's thinking that she'd like to manipulate it so that she can use the quad. equation on the denominator of the second equation.

 

Other suggestions?

[regentrude]

Daughter is stuck on a Physics problem.

 

She's got a system of equations with 2 unknowns:

 

Tcos(x) = 980

 

[Tsin(x)] = [6.67 * 10 ^ (-7)] / [1- 400tan(x) + 40,000tan^2(x) ]

 

x is in degrees, not radians

 

She's thinking that she'd like to manipulate it so that she can use the quad. equation on the denominator of the second equation.

 

Other suggestions?

 

Where did the equations originate? Looks like a force problem. The expression in the denominator is equal to [1- 200 tan(x)]^2, but how did it arise?

Sometimes going a step back may resolve issues, so it would help to know the actual problem. My gut feeling (conditioned by 12 years of teaching physics) tells me that there might possibly be something not quite right with the equation.

 

(As a side note: putting in numbers instead of symbols for the given quantities is a very bad technique she should not have been taught by her instructors)

[hsbeth]

Thanks for your help so far. The original problem was:

 

Two 100kg lead spheres are suspended from 100m long massless cables. The tops of the cables have been carefully anchored exactly 1m apart. What is the distance between the centers of the two spheres.

 

She drew a diagram of the system at equilibrium. Theta was defined as the angle between the cable and the ceiling. T is the tension force of the cable. She found that d (distance between the two spheres) was 1-2 [100tan(theta)].

 

She then drew a free body diagram and because the system is in equilibrium the sum of the forces is equal to 0. She got the equations: tcos(theta)=mg    and  Gm^2/d^2= t(sin)(theta)

 

In order to get it to a system of two equations with two unknowns, she used her previous equation for d and plugged it in.

 

If she can solve for theta she can just plug it into her equation for d?

 

 

[regentrude]

Thanks for your help so far. The original problem was:

 

Two 100kg lead spheres are suspended from 100m long massless cables. The tops of the cables have been carefully anchored exactly 1m apart. What is the distance between the centers of the two spheres.

 

She drew a diagram of the system at equilibrium. Theta was defined as the angle between the cable and the ceiling. T is the tension force of the cable. She found that d (distance between the two spheres) was 1-2 [100tan(theta)].

 

She then drew a free body diagram and because the system is in equilibrium the sum of the forces is equal to 0. She got the equations: tcos(theta)=mg    and  Gm^2/d^2= t(sin)(theta)

 

In order to get it to a system of two equations with two unknowns, she used her previous equation for d and plugged it in.

 

If she can solve for theta she can just plug it into her equation for d?

 

If theta is the angle between ceiling and cable, the trig functions should be reversed:

 

tsin(theta)=mg    and  Gm^2/d^2= t(cos)(theta)

 

Her expression for the distance d is wrong, because she made another mistake with her trig functions:

If theta is the angle between ceiling and cable, the distance between the spheres would be 1m - 2*100m*cos(theta). cos, not tan.

[Teachin'Mine]

And to think I'd answer 1m.  I can't imagine that the pull of the masses would be that significant, but I'm clearly not a physics person.  I'm assuming that the cable is centered on the sphere. 

[regentrude]

Still looks nasty though, even with the correct trig...I do not believe there actually is an analytical solution to the trig equation.

But we do know one thing: we know that the gravitational force will be very small, and thus the deviation from 1 m will be very small, or in other words, the angle theta will be very close to 90 degrees.

 

This is how I would approach the problem:

tsin(theta)=mg    and  Gm^2/d^2= t(cos)(theta) and d= x-2 Lcos (theta)

 

rearrange the first for t: t=mg/sin(theta))

 

and sub into the second gives:

 

mg cot (theta)= Gmm/(x-2L cos theta)^2

 

So far, everything is completely exact.

Now on the right hand side, the denominator contains a sum that is squared.

We are going to make an approximation: you know that the correction 2L cos theta will be very very tiny compared to the distance x=1m, so next to x=1m we can approximately ignore the correction of 2L cos theta.

In other words, the gravitational force will be pretty much the same as if the spheres were 1m apart.

 

This simplifies your equation:

mg cot (theta)= Gmm/(x)^2

or

theta = arccot(GM/(gx^2)

 

You should get an angle that is very close to 90 degrees.

You can then calculate the actual distance as

x-2L cos (theta)

and will see that this is very close to 1m, so the approximation we made was justified.

 

 

 

[hsbeth]

Okay, thanks. She does see those mistakes and has corrected them. They haven't covered "small angle approximation" yet. Really appreciate your help. She has kind of a bum prof. who doesn't keep office hours.

[regentrude]

Okay, thanks. She does see those mistakes and has corrected them. They haven't covered "small angle approximation" yet. Really appreciate your help. She has kind of a bum prof. who doesn't keep office hours.

 

I typed out my solution in my previous post and edited it a bit for clarity. Hope this helps. Turns out, there is an even simpler way than approximating the trig functions themselves (which, btw, she should have covered in calculus, not physics), but we still need to approximate.
 

 

Btw, typing the first sentence of her problem verbatim into a google search will give her several different sites where this problem is discussed.

[regentrude]

And to think I'd answer 1m.  I can't imagine that the pull of the masses would be that significant, but I'm clearly not a physics person.  I'm assuming that the cable is centered on the sphere. 

 

You would actually be very close to correct, since the force is tiny, and the deviation of the spheres due to their gravitational attraction will be tiny compared to their 1m distance. In fact, solving the problem requires us to use this very idea.

[hsbeth]

Thank you so much! I really appreciate your taking time to help with this.

[hsbeth]

When she attempts to calculate theta using the simplified equation she is getting a very tiny value, not anything close to 90 degrees?

She is getting something very close to 0.

[regentrude]

When she attempts to calculate theta using the simplified equation she is getting a very tiny value, not anything close to 90 degrees?

She is getting something very close to 0.

 

Then she made an error using her calculator.

 

You can do a simple estimate:

theta = arccot(GM/(gx^2)

if you insert numbers and round 9.8 to 10 to get order of magnitude:

 

theta= arccot ( 6.67E-12* 100kg/(10 m/s^2 * 1m^2))= arc cot 6.67 E -11

 

6.67 E-11 is approximately zero. The cotangent is cosine/sine, and that is zero wherever the cosine is zero, i.e. at 90 degrees. This is an estimate she should be performing mentally before using her calculator, to make sure the computed result makes sense.

 

Most calculators do not have a cotangent button. She very likely needs to input the fraction that equals the cotangent and invert it to get the tangent, and then compute the arc tan using the inverse function button.

My guess would be that she missed the step of inverting her expression and has actually calculated the arctan of her original expression, which would be close to zero degrees, since it is equal to 90 degrees minus theta.

 

She should be writing out all her expression with numbers put  in before punching them into the calculator, AND do a mental estimate without the calculator (as I showed above) to see whether her answer makes sense.

 

 

 

[hsbeth]

Got it, thanks again!

[MarkT]

Was this problem from a college class or AP?