<p>Physics questions: I have a couple of questions, hopefully someone can help. </p>
<p>1.) There are two 10 Kg masses attacached to an elevator. (Sorry I dont have the drawing i will try to explain.) Starting from the ceiling going down there is a rope, then a 10kg mass then another rope and the a 10kg mass. If the elevator accelerates upwards at 2m/s^2 what is the tension in each rope.</p>
<p>2.) My other question involves circular motion. First what is apparent weight when at the top of a hill or bottom. </p>
<p>mv^2 / r = Fg - Fn </p>
<p>I am not sure about that but I think that is for the top of the hill .</p>
<p>THanks for the help</p>
<p>T in the top rope = Ma and its mass = 10+10 and its acceleration = (g + 2m/s^2 upward) =20<em>11. Between the two masses the tension = mass *a which is just 10kg *11.<br>
Edit g= 10 not 9 so its 20</em>12</p>
<p>At top of the hill, it matters if you are over or under the track. If you are over, Fn (upward) - mg(downward)=mv^2/r
If you are under the track, then
-|Fn| - mg = mv^2/r so you can have 0 Fn and weight does all the pushing you down, and if you do it with the velocities and energies, you end up feeling 6mg's at the bottom.</p>
<p>I got 240N for the first one. Don't know if thats right though.</p>
<p>First T: T - mg = ma
T - 20(10) = 2(20)
T - 200 = 40
T = 240 N
Second T: T - mg = ma
T - 10(10) = 2(10)
T - 100 = 20
T = 120 N</p>
<p>Is that right?</p>
<p>k lemme see</p>
<p>(1)
second rope:</p>
<p>the lower mass has 2 force acting on it, those are mg and T2</p>
<p>T2- mg = ma
T2 = ma + mg = 120</p>
<p>first rope</p>
<p>the upper mass has 3 force acting on it, those are mg, T1 and T2</p>
<p>T1-T2 - mg= ma
T1 = ma + T2 +mg= 2(ma + mg) = 240</p>
<p>dono if it is right though, do u have the answer?</p>
<p>dont really understand wut u saying about second question</p>
<p>do we have to know apparent weight? We defintiely didn't learn about it</p>
<p>Is it two ropes or one rope? Is it like</p>
<p>ceiling
|
|
|
mass
|
|
mass</p>
<p>Here let me draw it</p>
<pre><code> Ceiling
</code></pre>
<p>Here let me draw it</p>
<pre><code> Ceiling
|
| T1
|
| 10 | a = 2 m / s^2 up
|
| T2
|
| 10 |
</code></pre>
<p>I thought it was mv^2 / r = in - out</p>
<p>If it's in a line, it's only one rope and theres only one tension.</p>
<p>junior....that would be true if the ropes weren't separated by the second weight....since the lower rope is only holding the mass of the bottom weight it's tension would be less then that of the top rope which has to account for the mass of both weights..I am also assuming that the masses of the ropes are negligible, otherwise there would be a different tension at each end of the ropes as well, but I don't see that as the question.
I believe it was already answered but all you would have to do to find each tension is first find the tension generated in the lower rope, then find the tension in the top rope without forgetting to add the mass of both blocks together to find the weight pulling down.</p>
<p>barce11 - Apparent weight is just the force the person/object experiences in that position.</p>
<p>What you wrote should be correct.</p>
<p>hm....no one confirmed this...did I do the second tension right?</p>
<p>Actually I have a very different solution.</p>
<p>Okay first we draw the force diagrams, right? Then we get the following two equations for each block:</p>
<p>top block: T - mg - T2 = ma (Where T2 = tension of second string)
Bottom block: T2 - mg = ma</p>
<p>Substitute 2nd equation into 1st:</p>
<p>T - 2mg - ma = ma
T = 2(10)(10) + (10)(2) + (10)(2)
T = 240N is the tension for the first rope</p>
<p>Then for the 2nd rope:</p>
<p>T2 = mg + ma
T2 = 10(10+2)
T2 = 120N</p>
<p>So yes, Earth-dragon is correct, I got the same answer.</p>