May 2nd 2008, 06:30 AM   #2
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 Originally Posted by Heavyarms2050 2. A meter stick can be balanced at the 49.7-cm mark when balanced on a fulcrum (pivot). When a 50.0-gram mass is attached at the 10.0-cm mark, the fulcrum must be moved to the 39.2-cm mark for balance. What is the mass of the meter stick in grams? Ans: 139 gm Yea, I'm clueless

I will help you with this question for now. Are you familiar with the principle of moments ?

If the meter stick is balanced at the 49.7cm mark, what does that tell you about the position of the centre of mass?

A good clear diagram always helps with a question like this, I have attached one.

first find the distance of the 10g block form the pivot.

39.2cm - 10cm = 29.2cm

and you need to find the distance of the centre of mass of the rules form the pivot.

49.7cm - 39.2cm = 10.5cm

using the principle of moment the anticlockwise moment causes by the mass is equal to the clockwise moment cause my the centre of mass about the pivot.

$\displaystyle 29.2cm \times 50 g = 10.5cm \times M$

solve the equation for M, you should get 139 grams.

Bobak
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Last edited by Bobak; May 2nd 2008 at 12:57 PM.

 May 2nd 2008, 12:54 PM #3 Junior Member   Join Date: May 2008 Posts: 3 $\displaystyle 29.2cm \times 10 g = 10.5cm \times M$ there was typo in your formula, it was suppose to be 50g not 10g. Thx a million though. This is actually getting urgent now, but can you guys/anybody PLEASE help me on these 4 problems by Monday.
May 3rd 2008, 11:18 AM   #4
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Posts: 5
 Originally Posted by Heavyarms2050 5. A thin spherical shell of mass 0.400 kg and diameter 0.200 m is filled with alcohol (rho = 806 kg/m3). It is released from rest at the bottom of a tank of water (rho = 1000 kg/m3). What is the acceleration of the alcohol spherical shell as it starts to rise to the surface? Ans:1.06 m/s^2 Yea, another clueless one

I believe you have to use archimedes principle for this question. Do you know what it is?

I will quote it for you.

 When a body is wholly or partly immersed in a fluid it experiences and upward force equal to the weight of the fluid displaced.

Firstly we must calculate the weight of the displaced water which. will be the volume of the sphere multiplied by the density of the water.

$\displaystyle Upthrust = 1000 Kg /m^3 \times \frac{4}{3} \pi (0.1)^3 m^3 \times 9.81 m/s^2$

now you must draw a free body force diagram

note the mass of the sphere is $\displaystyle 806 Kg /m^3 \times \frac{4}{3} \pi (0.1)^3 m^3 + 0.4 Kg$ this is the mass of the volume of the alcohol plus the mass of the shell

so the net force is $\displaystyle Upthrust - Weight$

using Newton's second law

$\displaystyle Mass \times Acceleration = Upthrust - Weight$
$\displaystyle ma= 1000 Kg /m^3 \times \frac{4}{3} \pi (0.1)^3 m^3 \times 9.81 m/s^2 - m \times 9.81 N$

so $\displaystyle a= \frac{1000 Kg /m^3 \times \frac{4}{3} \pi (0.1)^3 m^3 \times 9.81 m/s^2}{ 806 Kg /m^3 \times \frac{4}{3} \pi (0.1)^3 m^3 + 0.4 Kg } - 9.81 m/s^2$

giving $\displaystyle a= 1.07195 m/s^2$
or $\displaystyle a= 1.06 m/s^2$ quoting to 3 significant figures.

I hope that helps

Bobak
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Last edited by Bobak; May 3rd 2008 at 02:31 PM.

 May 3rd 2008, 02:21 PM #5 Junior Member   Join Date: May 2008 Posts: 3 $\displaystyle a= \frac{1000 Kg /m^3 \times \frac{4}{3} \pi (0.1)^3 m^3 \times 9.81 m/s^2}{ 806 Kg /m^3 \times \frac{4}{3} \pi (0.1)^3 m^3 + 0.4 Kg } - 9.81 m/s^2$ I'm not getting the answer using the formula you have given me for the 2nd Last edited by Heavyarms2050; May 3rd 2008 at 09:07 PM.
May 3rd 2008, 06:37 PM   #6

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 Originally Posted by Heavyarms2050 2. A meter stick can be balanced at the 49.7-cm mark when balanced on a fulcrum (pivot). When a 50.0-gram mass is attached at the 10.0-cm mark, the fulcrum must be moved to the 39.2-cm mark for balance. What is the mass of the meter stick in grams? Ans: 139 gm
You neglected to mention that you are using $\displaystyle g = 10 m/s^2$. As most of the world does not use this (unrealistic) assumption you should always mention it when asking for help.

We know that the meter stick is balanced at the 49.7 cm mark, so this is the location of the CM.

I am going to take the sum of the torques about the 39.2 cm mark and call a CCW torque positive.

$\displaystyle \sum \tau = (0.0500~kg)(0.292~m) - Mg(0.105~m) = 0$
where M is the mass of the meter stick.

Solve for M.

-Dan
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