An old streetcar rounds a flat corner of radius 10.5 m at 16 km/h. What angle with the vertical will be made by the loosely hanging and straps?

Answer

To solve the problem step by step, we will follow these instructions:

### Step 1: Convert the speed from km/h to m/s
The speed of the streetcar is given as 16 km/h. To convert this to meters per second (m/s), we use the conversion factor:
\[ \text{Speed in m/s} = \text{Speed in km/h} \times \frac{5}{18} \]
Calculating this gives:
\[ \text{Speed} = 16 \times \frac{5}{18} = \frac{80}{18} \approx 4.44 \, \text{m/s} \]

### Step 2: Identify the forces acting on the straps
When the streetcar rounds the corner, two main forces act on the straps:
1. The gravitational force (\(mg\)) acting downward.
2. The centripetal force (\(F_c\)) acting horizontally towards the center of the circular path.

### Step 3: Write the equations for the forces
The centripetal force required to keep the streetcar moving in a circle is given by:
\[ F_c = \frac{mv^2}{r} \]
where:
- \(m\) is the mass of the streetcar,
- \(v\) is the speed (which we calculated as approximately 4.44 m/s),
- \(r\) is the radius of the corner (10.5 m).

### Step 4: Set up the relationship between the forces
The angle \( \theta \) that the straps make with the vertical can be found using the relationship between the forces:
\[ mg \sin \theta = F_c \cos \theta \]
Dividing both sides by \(mg\):
\[ \tan \theta = \frac{F_c}{mg} \]

### Step 5: Substitute for \(F_c\)
Substituting \(F_c\) into the equation gives:
\[ \tan \theta = \frac{\frac{mv^2}{r}}{mg} = \frac{v^2}{rg} \]
Here, \(g\) is the acceleration due to gravity (approximately \(9.81 \, \text{m/s}^2\)).

### Step 6: Calculate \( \tan \theta \)
Now we can substitute the values:
\[ \tan \theta = \frac{(4.44)^2}{10.5 \times 9.81} \]
Calculating this gives:
\[ \tan \theta = \frac{19.7136}{103.905} \approx 0.189 \]

### Step 7: Find the angle \( \theta \)
Now we can find \( \theta \) using the arctangent function:
\[ \theta = \tan^{-1}(0.189) \approx 10.75^\circ \]
Rounding this gives:
\[ \theta \approx 11^\circ \]

### Conclusion
The angle with the vertical made by the loosely hanging straps is approximately \(11^\circ\).

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