In the ideal apparatus shown in the diagram, m1 = 2.0 kg. What is m2 if both masses are at rest? How about if both masses are moving at constant velocity?


Ideal aparatus is a pulley system with M1 on a 37 degree incline and m2 dangle on the other side of the incline

To find the value of m2, we first need to take into account the forces acting on the system.

1. When both masses are at rest:
In this case, the system is in equilibrium, which means the net force on both masses is zero. The forces acting on m1 are gravity (mg) acting vertically downward and the component of gravity parallel to the incline (mg*sin(37°)). The force acting on m2 is tension in the string (T).

Since the masses are at rest, the tension in the string must be equal to the component of gravity parallel to the incline. Mathematically, we can express this as:

T = mg*sin(37°)

Given that m1 = 2.0 kg, we can substitute this into our equation:

T = 2.0 kg * 9.8 m/s² * sin(37°)

Calculating this, we find T ≈ 11.92 N. Therefore, m2 = T/g.

2. When both masses are moving at constant velocity:
In this case, the system is experiencing dynamic equilibrium, which means the net force acting on each mass is zero. The forces acting on m1 are similar to the previous case (gravity and the component of gravity parallel to the incline) and are counteracted by the tension (T). However, now we need to consider the frictional force.

The frictional force opposes the motion on the inclined plane and is given by the equation:

frictional force = coefficient of friction * normal force

The normal force is the component of gravity perpendicular to the inclined plane, which is expressed as:

normal force = m1 * g * cos(37°)

Substituting this into our equation, we get:

frictional force = coefficient of friction * m1 * g * cos(37°)

Since the system is moving at constant velocity, the frictional force must be equal to the tension in the string. Now we can equate the two forces:

frictional force = T

coefficienf of friction * m1 * g * cos(37°) = T

Solving for T, we can find m2 as T/g.

Please provide any additional information if the coefficients of friction or other values are given in order to further calculate m2.

To determine the value of m2, we need to break down the problem and consider the forces acting on each mass separately.

First, let's consider the case where both masses are at rest. In order for the system to be in equilibrium, the net force acting on each mass must be zero.

For m1 on the incline:
- There is the force of gravity pulling it downward, which can be split into two components: mg * sin(37°) downhill along the incline, and mg * cos(37°) perpendicular to the incline.
- There is the normal force exerted by the incline, which is equal in magnitude and opposite in direction to the perpendicular component of the weight.
- Finally, there is the force of static friction between m1 and the incline, opposing the downward component of the weight.

For m2 hanging freely:
- There is only the force of gravity acting downward on m2.

Since both masses are at rest, the total force on each mass must be zero. Thus, we can equate the sum of the forces on each mass to zero.

For m1 on the incline:
mg * sin(37°) - μ * mg * cos(37°) = 0,
where μ is the coefficient of static friction.

For m2 hanging freely:
mg - T = 0,
where T is the tension in the string supporting m2.

Now, let's consider the case where both masses are moving at a constant velocity. In this case, the net force on each mass must still be zero, just as in the case when both masses are at rest.

For m1 on the incline:
mg * sin(37°) - μ * mg * cos(37°) = 0.

For m2 hanging freely:
mg - T = 0.

Solving these equations simultaneously will give us the values of m2 for both scenarios.

Note that we also need to consider the effect of the angle of the incline and the coefficient of static friction in order to find the exact values of m2.

Is there friction?