Two forces act on a 16 kg object. The first force has a magnitude of 68 N and is directed `24^(@)` north of east. The second force is 32 N, `48^(@)` north of west. What is the acceleration of the object resulting from the action of these two forces ?

Options

A. `1.6" m"//"s"^(2),5.5^(@)` north of east
B. `1.9" m"//"s"^(2),18^(@)` north of west
C. `2.4" m"//"s"^(2),34^(@)` north of east
D. `4.1" m"//"s"^(2),52^(@)` north of east

Answer

To solve the problem of finding the acceleration of a 16 kg object acted upon by two forces, we can follow these steps:

### Step 1: Identify the Forces and Their Components
We have two forces acting on the object:
1. **Force 1 (F1)**: 68 N at an angle of 24° north of east.
2. **Force 2 (F2)**: 32 N at an angle of 48° north of west.

We need to break these forces down into their x (east-west) and y (north-south) components.

**For Force 1 (F1)**:
- \( F_{1x} = 68 \cos(24°) \)
- \( F_{1y} = 68 \sin(24°) \)

**For Force 2 (F2)**:
- Since this force is directed north of west, we need to consider the negative x-direction:
- \( F_{2x} = -32 \cos(48°) \)
- \( F_{2y} = 32 \sin(48°) \)

### Step 2: Calculate the Components
Now we will calculate the components using a calculator.

**Calculating Force 1 components**:
- \( F_{1x} = 68 \cos(24°) \approx 68 \times 0.9135 \approx 62.1 \, N \)
- \( F_{1y} = 68 \sin(24°) \approx 68 \times 0.4067 \approx 27.6 \, N \)

**Calculating Force 2 components**:
- \( F_{2x} = -32 \cos(48°) \approx -32 \times 0.6691 \approx -21.4 \, N \)
- \( F_{2y} = 32 \sin(48°) \approx 32 \times 0.7431 \approx 23.7 \, N \)

### Step 3: Find the Resultant Force Components
Now, we can find the resultant force components by adding the corresponding components of both forces.

**Resultant Force in x-direction (R_x)**:
\[ R_x = F_{1x} + F_{2x} = 62.1 - 21.4 = 40.7 \, N \]

**Resultant Force in y-direction (R_y)**:
\[ R_y = F_{1y} + F_{2y} = 27.6 + 23.7 = 51.3 \, N \]

### Step 4: Calculate the Magnitude of the Resultant Force
The magnitude of the resultant force can be calculated using the Pythagorean theorem:
\[ R = \sqrt{R_x^2 + R_y^2} \]
\[ R = \sqrt{(40.7)^2 + (51.3)^2} \]
\[ R = \sqrt{1656.49 + 2632.69} \]
\[ R = \sqrt{4289.18} \approx 65.5 \, N \]

### Step 5: Determine the Direction of the Resultant Force
To find the direction (angle θ) of the resultant force:
\[ \tan(\theta) = \frac{R_y}{R_x} \]
\[ \theta = \tan^{-1}\left(\frac{51.3}{40.7}\right) \approx 52° \text{ north of east} \]

### Step 6: Calculate the Acceleration
Using Newton's second law, \( F = ma \), we can find the acceleration \( a \):
\[ a = \frac{R}{m} = \frac{65.5}{16} \approx 4.09 \, m/s^2 \]

### Final Answer
The acceleration of the object is approximately **4.1 m/s²** in the direction of **52° north of east**.

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