What Angle Causes the Fishline to Break?

[oneplusone]

Homework Statement

A 2 kg ball is attached to the bottom end of a length of fishline with a breaking strength of 44.5 N. The top end of the fishline is held stationary. The ball is released from rest with the line taut and horizontal (theta = 90 degrees). At what angle theta (measured w/vertical) will the fishline break?

Homework Equations

none i can think of

The Attempt at a Solution

$$F=\dfrac{mv^2}{l}$$ where l is the length of the wire. then from conservation:

$$mgh = \dfrac{1}{2}mv^2$$

$$mg(l \sin \theta) = \dfrac{1}{2} ( F\cdot l)$$

$$F = 2mg\sin \theta = 44.5$$

Which is incorrect as sin theta must be < 1.

 

[rock.freak667]

At the angle θ, what would be the components of the forces radially and tangentially and in what direction would the tension T act?


After that, how would you find the centripetal force in terms of these forces?

 

[oneplusone]

Sorry I still can't get it with your information. Any other hints/solution?

 

[Hollumber]

Account for the weight opposing tension force.

(At the bottom, Fnet = 2mg = Tension - mg(weight), so max tension would have to = 60N for the line to never break.)

 

[Doc Al]

Two comments:
- You have sine and cosine reversed; note that θ is measured from the vertical.
- What you call F is the radial component of the net force on the ball, not the tension in the string.

Draw a free body diagram of the forces on the ball.

 

[oneplusone]

Oops, is this correct now:

$$F=T-mg\cos\theta \implies T-mg\cos\theta = \dfrac{mv^2}{l}$$
Where l is the length of the line.

$$mgh = \dfrac{1}{2}mv^2\implies mg(l\cos\theta) = \dfrac{1}{2}(T-mg\cos\theta)\cdot l$$

$$2\cdot 9.8\cos\theta = \dfrac{1}{2}(44.5-2\cdot 9.8\cdot\cos\theta)$$

$$\theta = 40.82^{\circ}$$

Please correct if it's wrong.

 

[Hollumber]

It looks good to me