Urgent! Physics Help Requested

Test Preparation
1

<p>Here's the problem for all you smart people =] [please help btw]</p>

<p>A bell ringer decides to use a bowling ball to ring a BELL. He hangs the 7.3-kg ball from the end of a 2.0 m long rope. He attaches another rope to the ball to pull the ball back, and pulls it horizontally until the ball has moved .60 m away from the vertical. HOW MUCH FORCE MUST HE APPLY?</p>

<p>I got an answer but am not sure it's right. If you can--please, please, please--post your answer and [if you want to] an explanation.</p>

<p>P.S. -- I know the answer's in NewTons!</p>

2

<p>I'm definately shaky in my physics but I got 10.98 N.</p>

<p>Someone better check this. I'm not even sure if I understood the whole process of the problem without a proper diagram.</p>

3

<p>I got like 23 Newtons =/</p>

4

<p>What level physics is this from? Regular, Honors or AP B/C? I'm trying to work it and I think I'm overthinking it.</p>

<p>Also, when you say "moved .60 m away from the vertical", does this mean it moved horizontally .6m from the vertical line at the starting point?</p>

5

<p>Regular physics [not ap since school doesn't off it] problem -- first ball was hanging then moved .6 m to the right [so yes...that's what it means]</p>

6

<p>Can anyone figure it out?</p>

7

<p>It has been <em>decades</em> since I looked at physics problems, but here's a possible solution:</p>

<p>After you pull the ball 0.6 m from the vertical, you get a right-angled triangle with hypotenuse = 2m, and one side(horizontal) = 0.6m. The length of the third side (vertical) = sqrt( 2^2 - (0.6)^2) = 1.907m . </p>

<p>I think
(horizontal<em>force/ vertical</em>force) should = (horizontal<em>side / vertical</em>side)
Since gravitational force on 7.3 kg is 7.3 * (32/3.3) = 7.3 * 9.697= 70.8 newtons,</p>

<p>x / 70.8 = 0.6 / 1.907
x = (70.8)(0.6)/1.907 = 22.27 newtons</p>

<p>(assuming gravitational force exerted on 1 kg ball is (32/3.3) newtons)</p>

8

<p>i got 23 and u got 22.something..that's a good sign..probably just rounding error..thanks so much</p>

9

<p>but can anyone else verify this?</p>

10

<p>okay we're given that the string is 2 m long. and .6m away from vertical(i'm assuming .6m horizontally). it's 23.871 (no rounding) and significant digits make it 24 N.</p>

<p>oh and optimizerdad, it's not really necessary to know the length of the other side, just the angle that the rope forms(the angle between vertical and .6m away). then you can use the tangent of the angle to find the ratio between the horizontal and vertical components of the tension. you get:
tan(angle)=(Tension in x-direction)/mg.
|
V
mgtan(angle)=tension in x-direction
(7.3)(9.81)(tan(17.45...))=tension in x-direction = 23.871N</p>

11

<p>I agree with what you're saying...but do your math again</p>

<p>I think it's 22 N</p>

12

<p>I got 23 Newtons</p>

13

<p>Edit: search for a physics forum. You can PM me if you want a <a href="http://www"&gt;www&lt;/a>. of one, but I can't post it here.</p>

14

<p>I got 22.5N</p>