How to solve?

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1)A hot-air balloon of mass M is descending vertically with downward acceleration of magnitude a. How much mass (ballast) must be thrown out to give the balloon an upward acceleration of magnitude a? Assume that the upward force from the air (the lift) does not change because of the decrease in mass.

3)fig3
A worker wishes to pile a cone of sand onto a circular area in his yard. The radius of the circle is R, and no sand is to spill onto the surrounding area. If s is the static coefficient of friction between each layer of sand beneath it, show that the greatest volume of sand that can be stored in this manner is . (fig3)
 
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Answers and Replies

  • #2
Newton's second law

Hi

The figures didn't seem to work for your second question, but here's my attempt at an answer for the first.

First, choose downwards to be positive. Call the upwards force due to the hot air [itex] F_{up} [/itex], and the original mass of the balloon [itex] m_i [/itex]. Therefore

[tex] m_1 g - F_{up} = m_1 a[/tex]
[tex]m_1 = \frac{F_{up}}{g-a}[/tex].

([itex] g [/itex] is gravitational acceleration, therefore [itex] m_1 g [/itex] is the weight.)

Call the new mass after the ballast has thrown out [itex] m_2 [/itex]. The new acceleration is [itex] -a [/itex].

[tex]
m_2 g - F_{up} = - m_2 a[/tex]
[tex]
m_2 = \frac{F_{up}}{g+a}
[/tex].

The difference in the mass is [itex] m_1 - m_2 [/itex], which is

[tex]
m_1-m_2 = F_{up}\left( \frac{1}{g-a}-\frac{1}{g+a}\right) [/tex]

[tex]
m_1-m_2 = F_{up} \frac{2a}{g^2-a^2}
[/tex]
 

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