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I was watching an online video lecture about dynamics, and then I came across this brain teaser, and I've been thinking it over for a couple of hours but can't seem to find the solution. I hope someone will help me think about it. Here is the video: and the time for the question is 13:10.

So here is how I attempted this problem: For the person(p) the fnety=0... Fgp-Fnp=mpa, the scale will read the Fnp so Fnp=Fgp-mpa... ( the acceleration was found from a previous part in the video to be a=g(sin(theta)-cos(theta)*u) ) but I don't really think that's a problem that requires solving but thinking, so what can I do to help?

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Here is a hint, and a solution to go against.

If the person is accelerating downward (down slope), how much of the person's weight does the scale need to push back up? The tan box at the bottom of this answer has an equation which you should arrive at before you even start accounting for the slope.

Many people instinctively know the answer to this, but may find it hard to formalize. Your equation should agree with most people's instincts. The same thing happens when you go up or down on an elevator. I hope this helps you figure this out!

(I like to use the convention of using $-g$ instead of simply $g$ to show gravity is pulling things down. The $a_p$ term is the vertical component of the person's acceleration. The person's acceleration is also the vertical acceleration of your frame of reference, because the person stays on the scale without going through it.)

$ m_pa_p = -m_pg + F_n$

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  • $\begingroup$ I believe that the equation you wrote is the same as mine but with taking the positive direction to be up the slope. Right? However, I am still stuck. there is no any numbers to help me solve the equations, so all I know it will be less than real weight, but how can I know the actual value? $\endgroup$ – mariam.. May 29 '15 at 6:11
  • $\begingroup$ @mariam.. get a person, a scale, find a slope, and put in numbers. This is a general formula, so you can apply to to any situation where a person is on a scale, sliding down a hill. I think the professor here just wants you to think about it, not get numerical values. $\endgroup$ – PipperChip May 31 '15 at 14:13
  • $\begingroup$ I am sorry but I am still confused . $\endgroup$ – mariam.. Jun 1 '15 at 16:26
  • $\begingroup$ @mariam.. The general formula is for every case, every instance of someone riding down a slope on a scale. To get numerical values, you need the individual information of that case; your variables get replaced with numbers. You'll need the masses and speeds involved to get a number for your answer. $\endgroup$ – PipperChip Jun 1 '15 at 20:05

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