Two blocks are at rest on a frictionless, 20 degree incline. What are the tensions in the two strings if block 1= 5.3kg and block 2= 2.4kg?

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To determine the tensions in the two strings, we need to analyze the forces acting on each block separately.

Let's consider block 1, which has a mass of 5.3 kg. As the block is at rest on the incline, the forces acting on it are the gravitational force (mg) pulling the block downward, the normal force (N) from the inclined surface pushing it upward, and the tension in the string (T1) pulling it upwards.

The gravitational force pulling block 1 down the incline can be calculated using the formula: F = mg, where m is the mass of the block and g is the acceleration due to gravity (approximately 9.8 m/s²). Therefore, the gravitational force on block 1 is F1 = (5.3 kg)(9.8 m/s²).

Since the incline is frictionless, there is no frictional force acting on block 1. Therefore, the normal force N acting on block 1 is equal in magnitude and opposite in direction to the gravitational force, N = F1.

Block 1 is at an angle of 20 degrees with respect to the horizontal. Using the relationship between the normal force and the gravitational force, we can find the normal force acting on block 1.

N = F1⋅cos(θ), where θ is the angle of the incline. Therefore, N = F1⋅cos(20°).

Now, let's analyze block 2, which has a mass of 2.4 kg. Similar to block 1, the forces acting on block 2 are its gravitational force (mg) pulling it down, the normal force (N) pushing it up, and the tension in the string (T2) pulling it upwards.

The gravitational force on block 2 is F2 = (2.4 kg)(9.8 m/s²). The normal force N acting on block 2 is equal in magnitude but opposite in direction to the gravitational force.

Now we can analyze the equilibrium along the incline. Assuming the tension in the string T1 is greater than T2 (block 2 is farther up the incline), we can write equations for the vertical and horizontal components of the forces:

Vertical component: T1⋅sin(θ) - F1 - N = 0
Horizontal component: T2⋅sin(θ) - F2 = 0

Substituting F1, N, F2 into the equations, we get:

T1⋅sin(θ) - (5.3 kg)(9.8 m/s²) - (5.3 kg)(9.8 m/s²)⋅cos(20°) = 0
T2⋅sin(θ) - (2.4 kg)(9.8 m/s²) = 0

Solving these two equations will give us the values of T1 and T2, which are the tensions in the two strings.