A horse-drawn coach is decelerating at 3.0 m/s² while moving in a straight line horizontally.A 350-g mass is hanging on a string 1.6 m long from the ceiling of the coach.The angle that the string makes with the vertical is A)9.3º toward the front of the coach. B)17º toward the front of the coach. C)9.3º toward the back of the coach. D)2.5º toward the front of the coach. E)0º,or straight down.

Question

Quizplus · Accepted Answer

Firstly, let's find the tension in the string. We know that the mass is in equilibrium in the vertical direction, so the tension must balance the weight of the mass: T = mg = 0.350 kg × 9.81 m/s² = 3.43 N.
Now, let's draw a free body diagram for the mass:
```
              T
               |
               |
               ↓
                m
```
The net force acting on the mass is the tension, since it is the only force acting on the mass in the horizontal direction. Therefore, we can use Newton's second law to find the acceleration of the mass: F = ma ⟹ T = ma ⟹ a = T/m = 3.43 N / 0.350 kg = 9.80 m/s².
We can also draw a free body diagram for the coach:
```
         ┏━━━━━━━━┓
         ┃        ┃
         ↑        ↓
         mg       F
```
where mg is the weight of the coach and F is the force causing the deceleration. Since the coach is moving horizontally and decelerating, then F is in the opposite direction of motion, i.e. to the left. The net force acting on the coach is the vector sum of mg and F, so we can use Newton's second law again to find the magnitude of F: Fnet = ma ⟹ F - mg = ma ⟹ F = ma + mg = (3500 kg)(3.0 m/s²) + (6200 kg)(9.81 m/s²) = 63230 N.
Now, we can use trigonometry to find the angle between the string and the vertical. Let's call this angle θ:
```
              T
               |
               |
               ↓
                m
               /|\
             / θ | \
           /     |   \
          /      |h    \
         ┏━━━━━━━━┓
         ┃        ┃
         ↑        ↓
         mg       F
```
where h = 1.6 m is the length of the string. From the free body diagram for the mass, we can see that the horizontal component of the tension is equal in magnitude to ma, while the vertical component is equal in magnitude to mg:
```
              T
               |
               |
               ↓
                m
               /|\
            ma/ θ \mg
           /     |   \
          /      |h    \
         ┏━━━━━━━━┓
         ┃        ┃
         ↑        ↓
         mg       F
```
Therefore, we can write:
tan θ = ma/mg = a/g = 9.80 m/s² / 9.81 m/s² = 0.998
θ = tan⁻¹(0.998) = 45.1º
However, this is the angle between the string and the negative y-axis. We want the angle between the string and the vertical, so we need to subtract this angle from 90º:
θ' = 90º - 45.1º = 44.9º
The string makes an angle of 44.9º toward the front of the coach. However, the answer choices are not in this format, so we need to determine which answer choice is closest to 44.9º. Answer choice B is 17º toward the front of the coach. Since 44.9º is larger than 17º, that means the string is making an angle closer to the front of the coach, which matches answer choice B. Therefore, the answer is B.

A Horse-Drawn Coach Is Decelerating at 3