A force`F` has magnitude of `15N`. Direction of `F` is at `37^(@)` from nagative x-axis towards posititve y-axis. Represent F in terms of `hat i` and `hat J` .

Explanation

To represent the force \( F \) in terms of \( \hat{i} \) and \( \hat{j} \), we will follow these steps:

### Step 1: Understand the Direction of the Force
The force \( F \) has a magnitude of \( 15N \) and is directed at an angle of \( 37^\circ \) from the negative x-axis towards the positive y-axis.

### Step 2: Identify the Components of the Force
To find the components of the force in the x and y directions, we can use trigonometric functions. The x-component can be found using the cosine function, and the y-component can be found using the sine function.

### Step 3: Calculate the x-component
Since the angle is measured from the negative x-axis, the x-component will be negative:
\[ F_x = -F \cos(37^\circ) \]
Substituting the magnitude of the force:
\[ F_x = -15 \cos(37^\circ) \]

Using the cosine value:
\[ \cos(37^\circ) \approx \frac{4}{5} \]
Thus,
\[ F_x = -15 \times \frac{4}{5} = -12 \, \text{N} \]

### Step 4: Calculate the y-component
The y-component will be positive:
\[ F_y = F \sin(37^\circ) \]
Substituting the magnitude of the force:
\[ F_y = 15 \sin(37^\circ) \]

Using the sine value:
\[ \sin(37^\circ) \approx \frac{3}{5} \]
Thus,
\[ F_y = 15 \times \frac{3}{5} = 9 \, \text{N} \]

### Step 5: Write the Force in Vector Form
Now that we have both components, we can represent the force \( F \) in terms of \( \hat{i} \) and \( \hat{j} \):
\[ F = F_x \hat{i} + F_y \hat{j} = -12 \hat{i} + 9 \hat{j} \]

### Final Answer
The force \( F \) can be represented as:
\[ F = -12 \hat{i} + 9 \hat{j} \, \text{N} \]

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