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.

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"

To solve this problem, we need to analyze the forces acting on the system and apply the principles of static equilibrium.

Let's start by identifying the forces acting on each block:

1) Block on the left:
- Weight force (mg): This force acts vertically downward and can be calculated by multiplying the mass (48 Kg) by the acceleration due to gravity (9.8 m/s^2).
- Tension force (T): The tension in the rope is the same on both sides of the pulley, so it acts upwards on the left block.
- Normal force (N): This force acts perpendicular to the incline and counterbalances a portion of the weight force.

2) Block on the right:
- Weight force (mg): This force acts vertically downward and can be calculated by multiplying the mass (n3 Kg) by the acceleration due to gravity (9.8 m/s^2).
- Normal force (N): This force acts perpendicular to the incline and counterbalances a portion of the weight force.

To achieve static equilibrium, the sum of the forces in both the horizontal and vertical directions must be zero.

Let's analyze the forces in each direction:

1) In the vertical direction:
For the block on the left:
- mg - N = 0
Therefore, N = mg.

For the block on the right:
- mg - N = 0
Therefore, N = mg.

Since the normal force is counterbalancing the weight force for both blocks, the values of N for both blocks will be the same.

2) In the horizontal direction:
For the block on the left, the only horizontal force is the component of the weight force parallel to the incline, which can be calculated as:
- Force parallel = mg * sin(theta)

For the block on the right, there are no horizontal forces.

Since the system is in static equilibrium, the horizontal forces on both blocks must be equal and opposite.

Therefore, we have:
- Force parallel = 0

Taking the components of the weight force parallel to the incline:
- mg * sin(theta) = 0

Since the left incline is frictionless, there is no force parallel, which means that sin(theta) = 0.

Since sin(theta) = 0, the angle theta must be 0 degrees.

Therefore, the left incline is horizontal, and the right incline is vertical.

Now, let's answer the question:

The problem states that the left incline has a length of n1 meters, and the right incline has a length of n2 meters.

However, the problem does not provide sufficient information to determine the length of the right incline or the value of n3.

Therefore, we cannot determine the value of n3 based on the given information.