Two objects are connected by a massless string, as shown in the figure below. The incline and pulley are frictionless.

(a) Find the acceleration of the objects. (Assume that positive acceleration is to the right for m1 and downward for m2. Answer using m1 for m1, m2 for m2, g for the acceleration
due to gravity, and theta)
a =

Find the tension in the string.
T =

(b) Find the acceleration and tension for θ = 32° and m1 = m2 = 10 kg.
a =? m/s^2
T = ?N

This is kind of challenging without the figure but all you really need is F = m a and weight = mass * local gravity

The figure is just an inclined plane with two weights, both with a mass of 10kg but labeled M1 and M2 that are attached by a string. The pully is on the top point of the inclined plane and one weight is resting on the inclined plane with the other is hanging off the top point where the pully is. Does that make sense?

I really need help with this problem and I honestly have no clue how only knowing the force and weight will help find the tension if there is no friction.

T is tension

if the slope of the plane is 32 degrees
component of weight down plane =
10 (9.81) sin 32
which is trying to accelerate the mass down the slope
so if acceleration up slope is a
then
T - 98.1 sin 32 = 10 a
now the hanging mass
force down = mg - T

98.1 - T = 10 a

You now have two equations in T and a
add them to eliminate T
98.1 - 98.1 sin 32 = 20 a
solve for a
go back and solve for T

To find the acceleration of the objects, we can use Newton's second law of motion, which states that the net force on an object is equal to the product of its mass and acceleration (F = ma). We will consider both objects separately.

(a) For object m1:
1. Identify the forces acting on m1: the weight (mg) acting downward and the tension force (T) acting to the right.
2. Resolve the weight into its components: mg * sin(theta) acting perpendicular to the incline and mg * cos(theta) acting parallel to the incline.
3. Determine the net force on m1 along the incline:
- The force parallel to the incline is m1 * g * sin(theta), acting downhill.
- The force due to tension is T, acting uphill.
- Therefore, the net force on m1 is given by F_net = m1 * g * sin(theta) - T.
4. Apply Newton's second law to m1: m1 * a = F_net.
- Substitute the expression for F_net from step 3: m1 * a = m1 * g * sin(theta) - T.
5. Rearrange the equation to solve for the acceleration (a): a = g * sin(theta) - T / m1.

For object m2:
1. Identify the forces acting on m2: the weight (mg) acting downward and the tension force (T) acting upward.
2. Apply Newton's second law to m2: m2 * a = T - m2 * g.
- The force due to tension is T, acting upward.
- The force due to gravity is m2 * g, acting downward.
3. Rearrange the equation to solve for the acceleration (a): a = T / m2 - g.

To find the common value of acceleration (a) and the tension in the string (T), we need to solve the system of equations simultaneously:
a = g * sin(theta) - T / m1
a = T / m2 - g

Now let's substitute the given values into the equations to find the answers.

(b) For θ = 32° and m1 = m2 = 10 kg:
1. Substitute the values into the equations:
- For the acceleration equation: a = g * sin(32°) - T / 10.
- For the tension equation: a = T / 10 - g.
2. Equate the two equations: g * sin(32°) - T / 10 = T / 10 - g.
3. Simplify the equation: g * sin(32°) = 2T / 10.
- Multiply both sides by 10: 10g * sin(32°) = 2T.
4. Solve for T: T = 10g * sin(32°) / 2.
5. Substitute the value of T into either of the original equations to find the acceleration (a).

Therefore, the final answers for part (b) are:
- Acceleration (a) = the value obtained in step 5.
- Tension in the string (T) = the value obtained in step 4.