Make your coordinate system so that tension is positive for the object with the given mass and negative for the object with the unknown. So the x component for the first weight is going to be w1*cos(theta). We are not given the angle so we can rewrite this as n1/w1. This can be done similarly with the second weight which turn out to be n2/w2.
So now the summation across the x components must equal 0, since it is in equilibrium. So:
Tension-n1/w1 = 0, for the first block.
-Tension+n2/w = 0, for the second block.
Tension==n1/w1==n2/w2, since the tension will be the same for both block since pulleys change the direction of the force, not the magnitude.
Notice that gravity will cancel out so that we are only left with mass: Let mass2==n3;
n1/(48Kg*g)=n2*(n3*g), solve for n3.
n3=n2*48kg/n1.
Just plug the respective values for n2 and n1 and you won't get "OPPS Try Again"
Problem A5: In the figure above, the two blocks are on frictionless inclines and are in static equilibrium. The block on the left has a mass of 48 Kg, and the block on the right has a mass of n3 Kg. If the left incline has a length of n1 meters, and the right incline a length of n2 meters, what is n3? Assume that the mass of the rope is negligable and the pulley is frictionless.
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