A 3.6 kg mass hangs at one end of a rope that is attached to a support on a railroad car. When the car accelerates to the right, the rope makes an angle of 1.0° with the vertical. Find the acceleration of the car.

To find the acceleration of the car, we need to consider the forces acting on the hanging mass. As the car accelerates to the right, there are two forces acting on the hanging mass:

1. The weight of the mass, acting downward. The weight can be calculated as the mass multiplied by the acceleration due to gravity (g).
Weight = mass × g
Weight = 3.6 kg × 9.8 m/s^2 (approx. 9.81 m/s^2)
Weight = 35.28 N

2. The tension force in the rope, acting upward and at an angle with the vertical.

Let's consider the forces in the vertical direction:

Sum of forces in the vertical direction = Tension force - Weight

Since the rope makes an angle of 1.0° with the vertical, the vertical component of the tension force can be calculated as:
Vertical component of tension force = Tension force × sin(angle)
Vertical component of tension force = Tension force × sin(1.0°)

The tension force is the same in both vertical and horizontal directions. So, the horizontal component of the tension force can be calculated as:
Horizontal component of tension force = Tension force × cos(angle)
Horizontal component of tension force = Tension force × cos(1.0°)

In the horizontal direction, the only force acting is the horizontal component of the tension force.

Sum of forces in the horizontal direction = Horizontal component of tension force

Equating the forces in horizontal and vertical directions to their respective sums:

Horizontal component of tension force = Sum of forces in the horizontal direction
Tension force × cos(1.0°) = Sum of forces in the horizontal direction

Vertical component of tension force - Weight = Sum of forces in the vertical direction

Substituting the values:

Tension force × sin(1.0°) - Weight = 0 (since there is no vertical acceleration)

Now we can solve these two equations simultaneously to find the tension force (T).
T × cos(1.0°) = Sum of forces in the horizontal direction
T × sin(1.0°) - Weight = 0

Substituting the values:
T × cos(1.0°) = Sum of forces in the horizontal direction
T × sin(1.0°) - 35.28 N = 0

Now let's solve for T:

T × cos(1.0°) = T × sin(1.0°) - 35.28 N
T × (cos(1.0°) - sin(1.0°)) = 35.28 N

T = 35.28 N / (cos(1.0°) - sin(1.0°))
T ≈ 35.28 N / 0.99862953
T ≈ 35.316 N

Now, we can use the tension force (T) to find the acceleration of the car.
The tension force (T) provides the net force acting on the hanging mass:

Sum of forces in the horizontal direction = Tension force

Since the mass is hanging vertically, the horizontal component of the tension force provides the net force on the mass, which can be equated to the mass multiplied by the acceleration (a):

T = ma

Rearranging the equation:

a = T / m

a = 35.316 N / 3.6 kg
a ≈ 9.81 m/s^2

Therefore, the acceleration of the car is approximately 9.81 m/s^2.

To find the acceleration of the car, we can analyze the forces acting on the mass hanging from the rope.

First, let's draw a free-body diagram to visualize the forces:

```
Tension (T)
-----
| |
| m |
| |
Fg (-mg)
```

In this diagram:
- `Fg` represents the gravitational force acting on the mass, which is equal to its weight (-mg).
- `T` represents the tension in the rope.
- `m` represents the mass of the object that is hanging.
- `g` represents the acceleration due to gravity (approximately 9.8 m/s^2).

The tension in the rope can be resolved into two components: the vertical component (`Tsinθ`) and the horizontal component (`Tcosθ`), where `θ` is the angle made by the rope with the vertical.

Since the mass is in equilibrium in the vertical direction, the vertical component of the tension `Tsinθ` must balance the gravitational force `Fg`.

So we have the equation:

Tsinθ = Fg

From the given information, we know:
- Mass (m) = 3.6 kg
- Angle (θ) = 1.0°
- Acceleration due to gravity (g) ≈ 9.8 m/s^2

First, let's find the weight of the mass (Fg):
Fg = m * g = 3.6 kg * 9.8 m/s^2 = 35.28 N

Now we can rearrange the equation and solve for the tension (T):
Tsinθ = Fg
T = Fg / sinθ

We can substitute the known values into the equation:
T = 35.28 N / sin(1.0°)

Now we have the tension in the rope (T). However, we also need to consider the horizontal component of the tension to find the acceleration of the car.

The horizontal component of the tension (`Tcosθ`) causes the car to accelerate. According to Newton's second law, the net force acting on the mass is equal to the mass times its acceleration.

So we can write the following equation:

Tcosθ = m * a

Substituting the known values:
Tcosθ = 3.6 kg * a

Finally, we can rearrange the equation and solve for the acceleration (a):
a = Tcosθ / m

Substitute the values of T and θ into the equation:
a = (35.28 N / sin(1.0°)) * cos(1.0°) / 3.6 kg

Now you can calculate the acceleration of the car.

The vertical component of the rope tension equals the weight (M g) and the horizontal component equals the acceleration (M a). The ratio of horizontal to vertical components is the tangent of the tilt angle with respect to vertical.

Use those facts to solve for the acceleration, a

"M a" is the accelerating force, not the acceleration itself.

The rest of my previous answer is correct.

tan 1.0° = 0.01746 = Ma/Mg = a/g

Solve for a. g = 9.81 m/s^2

The 3.6 kg mass M does not matter. It cancels out.