a ball of mass m1 and a block of mass m2 are attached by a lightweight cord that passes over a frictionless pulley of negligible mass. the block lies on a frictionless incline of angle theta.

(a) draw the free body diagrams
(b) express newton's second law in component form
(c) find the magnitude of the acceleration of the two objects and the tension in the cord
(d) find the angle theta of zero acceleration for m2 = 2m1
(e) what's the magnitude of the tension at the condition of (d)
the gravitational acceleration is g.

a ball of mass m1 and a block of mass m2 are attached by a lightweight cord that passes over a frictionless pulley of negligible mass. the block lies on a frictionless incline of angle theta.

(a) draw the free body diagrams
(b) express newton's second law in component form
(c) find the magnitude of the acceleration of the two objects and the tension in the cord
(d) find the angle theta of zero acceleration for m2 = 2m1
(e) what's the magnitude of the tension at the condition of (d)
the gravitational acceleration is g.

(a) Free Body Diagrams:

For the ball of mass m1:
- There is a force due to gravity acting downwards (mg).
- There is tension in the cord acting upwards.
- There is no force due to friction because the pulley is frictionless.

For the block of mass m2:
- There is a force due to gravity acting downwards (mg).
- There is a normal force perpendicular to the incline.
- There is a force due to friction acting upwards along the incline.

(b) Newton's Second Law in Component Form:

For the ball of mass m1:
- In the vertical direction: T - m1g = m1a (since the ball can only move vertically)
- There is no horizontal component of motion for the ball, so no equation is needed.

For the block of mass m2:
- In the vertical direction: N - m2g - F_friction = m2a (since the block moves vertically)
- In the horizontal direction: F_friction = m2gsin(theta) (since the block moves horizontally)

(c) Magnitude of Acceleration and Tension:

Solving the equations from (b) simultaneously, we can eliminate the normal force (N) and solve for the acceleration (a):
m2gsin(theta) - m2g - m2a = m2a
m2g(1 + sin(theta)) = (2m2 + m1)a

Using this acceleration value, we can find the tension (T):
T = m1g + m1a

(d) Angle Theta of Zero Acceleration for m2 = 2m1:

Setting the acceleration (a) to zero, we can find the angle theta where the system is in equilibrium:
0 = m2g(1 + sin(theta)) - (2m2 + m1)a
0 = 2mg(1 + sin(theta)) - 3ma
3ma = 2mg(1 + sin(theta))
3a = 2g(1 + sin(theta))
3a/2g = 1 + sin(theta)
1 - 3a/2g = sin(theta)
theta = arcsin(1 - 3a/2g)

(e) Magnitude of Tension at Condition of (d):

Substituting the angle theta from (d) into the tension equation:
T = m1g + m1a
T = m1g + m1(0)
T = m1g

Therefore, the magnitude of tension at the condition of (d) is equal to m1g.