# Forces revisted

1. Oct 23, 2007

I am trying to help a friend with this webassign problem:

Jean, who likes physics experiments, dangles her watch from a thin piece of string while the jetliner she is in takes off from Dulles Airport . She notices that the string makes an angle of 23° with respect to the vertical while the aircraft accelerates for takeoff, which takes about 18 seconds. Estimate the takeoff speed of the aircraft.

I just need a kick in the rear to get me started! I know that there is Tension and weight acting on the watch. I know that average a=v/t so average v=at and I know that F=ma
But I am having trouble putting it all together since I do not have any mass or other numbers besides 18s.

Anyone got a hint for me?

Thanks,
Casey

2. Oct 23, 2007

### andrevdh

The vertical component of the tension supports the weight of the watch while the horizontal component provides the acceleration force.

3. Oct 23, 2007

Right, I am just not sure what to do without T or mg.....hmmmm

4. Oct 23, 2007

### odie5533

Try breaking the Tension up into it's components and apply Newton's 2nd law.

5. Oct 23, 2007

Obviously I am missing something big here. I do not have T; if I break it into components, I just get T_x=Tsin23 and T_y=cos23.... so by Newton's 2nd get $$\sum F_x=ma$$ so $$T\sin23=ma$$

...I am missing how to use 18 seconds....
....and I don't see how I could cancel anything..arrrgggggghhh

What am I overlooking?
Casey

6. Oct 23, 2007

### odie5533

Try applying Newton's 2nd law to $$\sum F_{y}$$ too.

This is actually a neat and useful solution, and simple enough that once you solve it you'll be able to remember it always :)

Last edited: Oct 23, 2007
7. Oct 23, 2007

So...$$\sum F_y=0$$ -->$$T_y-mg=0$$ --->$$T_y=mg=T\cos23$$ so...hey....$$m=\frac{T\cos23}{g}$$....looks like T will cancel if I plug that in.

I can't believe that they would give this to a high school student as the first of 15 problems....way to discourage the crap out of a bunch of kids who can't stand physics/math anyway.

I really think they should ease into the assignment a little more. Get these kids brains warmed up.

Christ...I would have given up if it were assigned to me.

thanks for the help,
Casey

8. Oct 24, 2007

### andrevdh

the horizontal component of the tension will be

$$T_H = T\sin(\theta)$$

and the vertical

$$T_V = T\cos(\theta)$$

but the horizontal component need to accelerate the watch

$$T_H = ma$$

and the vertical need to cancel the weight

$$T_V = mg$$

so

$$\frac{T_H}{T_V} = \frac{ma}{mg} = \frac{T\sin(\theta)}{T\cos(\theta)} = \tan(\theta)$$

giving

$$a = g \ tan(\theta)$$

9. Oct 24, 2007

### odie5533

This can also be solved geometrically:

10. Oct 24, 2007