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im stumped on this question please help! |
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| Oct1-05, 02:07 PM | #1 |
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im stumped on this question please help!
hey guys i am stumped on this question:
mary dangles her watch from a thin piece of string, while burning rubber on her supercharged moped. she notices that the string makes an angle of 15degrees with respect to the vertical while the moped accelerates to maximum speed, which takes about 18s. what is the maximum speed of mary's moped? i am really stuck on this question and i need to show my work too. can anyone please help? thanks. |
| Oct1-05, 02:16 PM | #2 |
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What, exactly, have you done so far?
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| Oct1-05, 02:19 PM | #3 |
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All you have to do is work out the acceleration on the watch/moped.
The watch is at an angle of 15 deg. Draw a diagram of the forces on the watch. What are the only forces acting ? From this, you should be able to get the acceleration. |
| Oct1-05, 02:38 PM | #4 |
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im stumped on this question please help!
i cant get anything, i just keep staring at the diagram but i cant figure it out, how do i figure out the acceleration by using the angle? do i brake it into the X and Y components? then what?
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| Oct1-05, 05:35 PM | #5 |
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is there no one that can help me?
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| Oct1-05, 05:37 PM | #6 |
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| Oct1-05, 05:37 PM | #7 |
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Have you got the diagram ?
What forces have you got labelled on it ? |
| Oct1-05, 05:38 PM | #8 |
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Ah, already answered.
So, now resolve the Tension, Ft, into horizontal and vertical components. |
| Oct1-05, 05:45 PM | #9 |
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how? it only gives me the angle and the time, no forces or mass of the watch. |
| Oct1-05, 05:48 PM | #10 |
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If you know the angle the tension is acting at, can't you just then resolve the tension, call it Ft, into horizontal and vertical components using that angle ?
Take the mass of the watch as m. |
| Oct1-05, 05:50 PM | #11 |
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Ftcos15= Ftx Ftsin15= Fty and then Ftx= m(ax)? then what do i substitute? the time comes from: accelerates to maximum speed, which takes about 18s. what is the maximum speed of mary's moped? the question gives it to me |
| Oct1-05, 05:56 PM | #12 |
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Almost there.
Ft.cos15 is actually the vertical component. You can equate this to the only other vertical force, the mass of the watch, hence you can get an expression for Ft. |
| Oct1-05, 05:59 PM | #13 |
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| Oct1-05, 06:03 PM | #14 |
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I'll upload a sketch.
The masss, m, should cancel out., |
| Oct1-05, 06:06 PM | #15 |
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See attachment.
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| Oct1-05, 06:06 PM | #16 |
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What is your expression for Ft ?
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| Oct1-05, 06:15 PM | #17 |
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