on a meter stick a fulcrum is placed at 50cm mark. Because of this fulcrum the meter stick can rotate about it. Then a 3kg mass is hung from the meter stick at 10cm mark. Where should a 5kg mass be placed on the meter stick so that the meter stick is in equilibrium?

Question

anonymous · Answer

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the system is in rotational equilibrium about the fulcrum
so, 
$$\Sigma\tau=0$$

anonymous · Answer

i understand it at equilibrium but what is the location on the meter stick of the 2nd mass??

anonymous · Answer

find "x"

anonymous · Answer

set up the equation first.. clockwise torques negative and counter cllockwise positive

anonymous · Answer

$$\Sigma=(3)(10)-(5)(x)=0$$

anonymous · Answer

nope... distances from the point of rotation...
3(50-10)-(5)(x-50)=0

anonymous · Answer

okay got it the answer is 74cm??

anonymous · Answer

sounds reasonable.. x-50 should be less than (50-10)

anonymous · Answer

got it thanks!

anonymous · Answer

a mass 20kg starts from rest at the top of the an incline. This mass reaches the bottom of the incline(30 degrees) with a speed of 9.90 m/s. What is the height of the incline ? What is the length of the incline? friction less surface. 

I got 5m as the height how do i get the length?