# String Problems?

1. Aug 17, 2009

### MorganJ

A 0.2-kg mass is hung from a string in an airplane awaiting take off. As the plane accelerates the string makes an angle of 26.7 degrees with the vertical, to the nearest hundredth of a m/s2 what is the acceleration of the plane?

I tried doing mgsin(theta)=ma.

3. (0.2)(9.8)(sin26.7)=(0.2)(a)
0.8806652373=0.2a
08806652323/0.2=a
a=4.4???

A fisherman yanks a fish straight up out of the water with an acceleration of 2 m/s^2 using very light fishing line that has a "test" value of 8 pounds. The fisherman, unfortunately, loses the fish as the line snaps. To the nearest tenth of a kilogram what is the minimum mass of the fish?

I have no idea how to solve this equation. I tried doing F=ma but to no avail. Can you help me?

2. Aug 17, 2009

### Staff: Mentor

How did you arrive at this equation? Hint: Analyze the forces on the mass and apply Newton's 2nd law to both the horizontal and vertical directions.

Newton's 2nd law is the way to go. What forces act on the fish?

3. Aug 17, 2009

### LBloom

1) Draw a free body diagram and you'll see that since they are giving you the angle with respect to the vertical, so you need to be careful when analyzing how you find the vertical and horizontal components.

2)Remember that pounds and kilograms do not measure the same thing. Pounds is a unit of force and is therefore equivalent to newtons in the metric system. If they tell you the most pounds the fishline can handle, they're telling you how much force it can handle, so you need to do some conversions from the English system to the Metric system.

4. Aug 18, 2009

### MorganJ

1) Is it force applied, tension, and normal force coupled together? I'm a little confused on how to even get the proper equation to solve the problem.

2)Same question.

5. Aug 18, 2009

### Staff: Mentor

In each problem the only forces acting on the mass (or fish) are string tension and weight. (There's no normal force.)

In the first problem, since the string is at an angle the tension acts at an angle. So you'll end up with two equations.

6. Aug 18, 2009

### MorganJ

2. So it wouldn't be T= M(g+a)

7. Aug 18, 2009

### LBloom

for the second question, the tension would be T=M(g+a)
as for the first question, the vertical component of the tension equals the weight of the mass since the mass is not accelerating in the y direction, so if you can find the equation for the vertical component of tension, you can find the tension itself.

8. Aug 18, 2009

### Staff: Mentor

That's correct.

9. Aug 18, 2009

### MorganJ

1. It is asking for acceleration so I don't understand how the vertical component will get me to the acceleration.

Would it be mgsin(theta) or...?

10. Aug 18, 2009

### LBloom

1)mgsin(theta) will only give you the vertical component of tension if the angle is given with respect to the horizontal. If you want, you can find the horizontal angle by finding the complementary angle of the vertical angle they give you and then use mgsin(theta) instead of mgcos(theta) with the angle they give you. however, you dont have to and w/ the equations I write out i'll be using the angle with respect to the vertical.
2) Ty=Tcos(theta)=mg
using this equation, since you know theta,m, and g, you can find the tension in the string. From there Ft=ma will give you the rest

If you prefer, after finding T, you can find the horizontal acceleration and use pythagoreams theorem instead.

11. Aug 18, 2009

### Staff: Mentor

You have two unknowns--acceleration and tension--so you need two equations. As LBloom stated, using an equation for vertical force components will give you the tension; the horizontal force components will then give you the acceleration.

Do not try to do it "in your head". Write the two force equations.

12. Aug 19, 2009

### MorganJ

Ok. SoI should find the complementary angle of 26.7 degrees. Then, I should set up two equations: Ty=Tcos(theta)=mg and once I solve for Ty then I do Ft=ma, which will give me acceleration?

13. Aug 19, 2009

### Staff: Mentor

Not unless you want to, for some reason.
Good, your first equation is: Tcosθ = mg.
Solve for T.
I don't know what Ft stands for, but apply F = ma for the horizontal component of force. (If the vertical component of tension is Tcosθ, what's the horizontal component?)

14. Aug 19, 2009

### MorganJ

The horizontal component is Tsin(theta)?

15. Aug 19, 2009

### LBloom

if angle is given w/ respect to horizontal
Tx=Tcos(theta)
Ty=Tsin(theta)

if angle is given with respect to the vertical
Tx=Tsin(theta)
Ty=Tcos(theta)

You only need to find the complementary angle, and therefore what the angle is with respect to the horizontal, if you want to use the first set of equations. If you use the second set, you won't need to find the complementary angle.

16. Aug 19, 2009

### Staff: Mentor

Exactly.

17. Aug 19, 2009

### MorganJ

So since it is in respect to the vertical, I use the second set of equations. So I do T(cos(theta)=mg and T(sin(theta)=mg then do T=ma? I am a little lost.

18. Aug 19, 2009

### Staff: Mentor

Yes.
No.

In all cases, you are applying Newton's 2nd law: ΣF = ma.

The acceleration is purely horizontal, so the vertical acceleration is 0.

The vertical equation:
ΣF = ma
Tcosθ - mg = 0 (or: Tcosθ = mg)
Since θ and m are given, you can solve for the tension T.

The horizontal equation:
ΣF = ma
Tsinθ = ma
Since you already found T, you can solve for the acceleration a.

19. Aug 19, 2009

### MorganJ

Ahhh I got it! Yes! Thank you sooo much for taking time out out of your day to help me, Doc Al and LBloom!