A mover has to move a heavy sofa of mass 43 kg to the second floor of the house. As he pulls on the rope tied to the sofa he makes sure that the rope is parallel to the surface of the ramp, which is at 30.0° to the horizontal. If the coefficient of kinetic friction between the sofa and the ramp is 0.350, and the sofa has an acceleration of 0.700 m/s2, find the tension in the rope in newtons.

We can solve for the tension in the rope using Newton's second law and breaking the forces down into components parallel and perpendicular to the inclined plane.

### Step 1: Free Body Diagram
The forces acting on the sofa are:
1. **Tension force (T)** pulling the sofa up the ramp.
2. **Gravitational force (mg)** acting vertically downward.
3. **Normal force (N)** perpendicular to the ramp.
4. **Frictional force (Fₖ)** opposing the motion of the sofa, acting down the ramp.

### Step 2: Break Forces into Components
- The gravitational force can be split into:
  - **Parallel component**: \( F_{\text{gravity, parallel}} = mg \sin \theta \)
  - **Perpendicular component**: \( F_{\text{gravity, perpendicular}} = mg \cos \theta \)

Where \( \theta = 30^\circ \), \( m = 43 \, \text{kg} \), and \( g = 9.81 \, \text{m/s}^2 \).

### Step 3: Normal Force
The normal force \( N \) balances the perpendicular component of the gravitational force:
\[
N = mg \cos \theta
\]
\[
N = (43 \, \text{kg})(9.81 \, \text{m/s}^2) \cos 30^\circ = 364.83 \, \text{N} \cdot 0.866 = 316 \, \text{N}
\]

### Step 4: Frictional Force
The kinetic friction force is given by:
\[
F_k = \mu_k N
\]
\[
F_k = 0.350 \times 316 \, \text{N} = 110.6 \, \text{N}
\]

### Step 5: Net Force
The net force acting parallel to the incline is:
\[
F_{\text{net}} = T - F_{\text{gravity, parallel}} - F_k
\]
Where the parallel component of the gravitational force is:
\[
F_{\text{gravity, parallel}} = mg \sin \theta
\]
\[
F_{\text{gravity, parallel}} = (43 \, \text{kg})(9.81 \, \text{m/s}^2) \sin 30^\circ = 43 \times 9.81 \times 0.5 = 210.015 \, \text{N}
\]

### Step 6: Newton's Second Law
We know that the net force also equals \( ma \), where \( a = 0.700 \, \text{m/s}^2 \). Thus:
\[
F_{\text{net}} = ma = (43 \, \text{kg})(0.700 \, \text{m/s}^2) = 30.1 \, \text{N}
\]

### Step 7: Solve for Tension
Now, we can solve for the tension \( T \):
\[
F_{\text{net}} = T - F_{\text{gravity, parallel}} - F_k
\]
\[
30.1 \, \text{N} = T - 210.015 \, \text{N} - 110.6 \, \text{N}
\]
\[
T = 30.1 \, \text{N} + 210.015 \, \text{N} + 110.6 \, \text{N} = 350.7 \, \text{N}
\]

### Final Answer:
The tension in the rope is \( T = 350.7 \, \text{N} \).

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Would you like a breakdown of any particular part of the solution or further details?

Here are 5 relative questions to deepen your understanding:
1. How does the coefficient of kinetic friction affect the tension in the rope?
2. What would happen to the tension if the incline angle increased?
3. How would a different acceleration impact the overall tension?
4. What role does the normal force play in calculating friction?
5. How would the scenario change if the ramp were frictionless?

**Tip**: When dealing with inclined planes, always break forces into parallel and perpendicular components to simplify calculations!

A mover has to move a heavy sofa of mass 43 kg to the second floor of the house. As he pulls on the rope tied to the sofa he makes sure that the rope is parallel to the surface of the ramp, which is at 30.0° to the horizontal. If the coefficient of kinetic friction between the sofa and the ramp is 0.350, and the sofa has an acceleration of 0.700 m/s², find the tension in the rope in newtons.

Learn how to calculate the tension in a rope as it moves a 43 kg sofa up a ramp inclined at 30° with a coefficient of kinetic friction of 0.350. This physics problem involves Newton's second law, inclined plane dynamics, and frictional forces. Suitable for Grades 10-12 students.

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