Consider a version of the atwood machine in which the masses are frictionless incline. The mass sliding on incline 1, m1, is 1.5 kg, and the angle of this incline is 62 deg. If the mass on the second incline, m2, is 2.5 kg, what is the angle 2 so that the system does not accelerate?
Did you leave out some words in the first sentence? The masses are not the incline.
There will be no system acceleration if the components of both weights in their respective directions down the incline slope are equal.
That will happen if m1*sin62 = m2*sinX
Solve for angle X
sinX = 0.5298
X = ?
To solve this problem, we need to analyze the forces acting on each mass and set up equations based on the conditions of the system not accelerating.
Let's consider the forces acting on mass m1, which is sliding down the incline:
1. The weight of m1 (W1) can be calculated as W1 = m1 * g, where g is the acceleration due to gravity.
2. The normal force (N1) acting perpendicular to the incline can be determined by N1 = m1 * g * cos(θ1), where θ1 is the angle of incline 1 in radians.
3. The component of the weight acting parallel to the incline (W1_parallel) is W1_parallel = m1 * g * sin(θ1).
Similarly, let's consider the forces acting on mass m2:
1. The weight of m2 (W2) is W2 = m2 * g.
2. The normal force (N2) acting perpendicular to the incline can be determined by N2 = m2 * g * cos(θ2), where θ2 is the angle of incline 2 in radians.
3. The component of the weight acting parallel to the incline (W2_parallel) is W2_parallel = m2 * g * sin(θ2).
Now, since the system does not accelerate, the net force acting on each mass must be zero. In other words, the sum of the forces along the incline must be balanced by the friction force.
For mass m1:
W1_parallel - μ * N1 = 0, where μ is the coefficient of kinetic friction between m1 and incline 1.
For mass m2:
W2_parallel + μ * N2 = 0, where μ is the coefficient of kinetic friction between m2 and incline 2.
Since the problem states that the masses are frictionless, we can conclude that the friction forces are negligible (μ = 0). Therefore, the equations become:
W1_parallel = 0, and W2_parallel = 0.
Substituting the expressions for the components of the weights, we get:
m1 * g * sin(θ1) = 0, and m2 * g * sin(θ2) = 0.
Since sin(θ1) ≠ 0 (θ1 ≠ 0°) and sin(θ2) ≠ 0 (θ2 ≠ 0°) in order for the system to be in equilibrium, we can conclude that both angles must be zero. In other words, both inclines must be horizontal (flat).
Therefore, the answer is that angle 2 (θ2) would be zero degrees.