A child is trying to pull a sled with mass 1.65 kg by a rope. He is able to pull with tension 12.3 Newtons. Because of the friction between the ground and the sled, with a coefficient of static friction *μs*​=1 (unrealistic), he needs to pull the rope off the ground at some *minimum* angle θ *t*o get the sled to move.
Find sin⁡θ

To solve this problem, we need to find the minimum angle \( \theta \) at which the child needs to pull the sled to overcome static friction. We are given the following:

- Mass of the sled, \( m = 1.65 \, \text{kg} \)
- Tension in the rope, \( T = 12.3 \, \text{N} \)
- Coefficient of static friction, \( \mu_s = 1 \)

### Step 1: Forces Acting on the Sled
When the child pulls the sled, two major forces affect its movement:
1. **Gravitational Force**: The weight of the sled is \( W = mg \), where \( g = 9.8 \, \text{m/s}^2 \).
   
   So,
   \[
   W = 1.65 \, \text{kg} \times 9.8 \, \text{m/s}^2 = 16.17 \, \text{N}
   \]
   
2. **Static Friction Force**: The maximum static friction force is \( F_{\text{friction}} = \mu_s N \), where \( N \) is the normal force.

### Step 2: Components of the Tension Force
The tension force \( T \) is applied at an angle \( \theta \), so it has two components:
- **Horizontal Component**: \( T_x = T \cos \theta \)
- **Vertical Component**: \( T_y = T \sin \theta \)

### Step 3: Normal Force
The normal force \( N \) is the force exerted by the ground on the sled. It's affected by both the weight of the sled and the vertical component of the tension force:
\[
N = W - T_y = W - T \sin \theta
\]
Substitute the values:
\[
N = 16.17 \, \text{N} - 12.3 \sin \theta
\]

### Step 4: Condition for Overcoming Static Friction
For the sled to start moving, the horizontal component of the tension force \( T_x \) must be equal to the maximum static friction force \( F_{\text{friction}} \):
\[
T_x = F_{\text{friction}} = \mu_s N
\]
Substitute \( T_x = T \cos \theta \) and \( F_{\text{friction}} = \mu_s N \):
\[
T \cos \theta = \mu_s (W - T \sin \theta)
\]
Since \( \mu_s = 1 \), this simplifies to:
\[
T \cos \theta = W - T \sin \theta
\]

### Step 5: Solve for \( \sin \theta \)
Now we solve the equation \( T \cos \theta = W - T \sin \theta \). First, rearrange it as:
\[
T \cos \theta + T \sin \theta = W
\]
Factor out \( T \):
\[
T (\cos \theta + \sin \theta) = W
\]
Solve for \( \cos \theta + \sin \theta \):
\[
\cos \theta + \sin \theta = \frac{W}{T} = \frac{16.17}{12.3} \approx 1.315
\]
Now, we use the identity \( \sin \theta + \cos \theta = \sqrt{2} \sin \left( \theta + \frac{\pi}{4} \right) \) to relate this expression, but since the angle requires finding components, we will work to numerically simplify.



A child is trying to pull a sled with mass 1.65 kg by a rope. He is able to pull with tension 12.3 Newtons. Because of the friction between the ground and the sled, with a coefficient of static friction μs=1, he needs to pull the rope off the ground at some minimum angle θ to get the sled to move. Find sin⁡θ.

In this problem, a child pulls a sled with mass 1.65 kg using 12.3 N of force at an angle θ to overcome static friction (μs=1). Learn how to calculate sinθ, break down the forces acting on the sled, and use trigonometric equations to find the minimum angle for movement. Suitable for high school physics students.

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