A 12.7 kg package is pulled with a 28.5 N force at a [tex]60.3^{\circ}[/tex] angle, across the ground where the coefficient of friction ([tex]\mu_k[/tex]) is equal to 0.0802.

What is the force of friction on the package?
[tex]F_f = \, ? \, \text{N}[/tex]



Answer :

To solve this problem correctly, we need to follow several important steps that involve the concepts of forces, angles, and friction. Let's break it down step by step:

### 1. Understanding the Scenario
We have a package with a mass of 12.7 kg that is being pulled by a force of 28.5 N at an angle of 60.3° above the horizontal. The coefficient of kinetic friction (µₖ) between the package and the ground is 0.0802. We need to find the force of friction acting on the package.

### 2. Converting Angle to Radians
First, we need to convert the angle from degrees to radians for our calculations, since trigonometric functions in certain operations require the angle in radians:
[tex]\[ 60.3^\circ = 1.0524335389525807 \, \text{radians} \][/tex]

### 3. Calculating the Normal Force
The normal force is the perpendicular force exerted by the ground on the package. It can be affected by both the weight of the package and the vertical component of the pulling force.

The forces acting in the vertical direction include:
- Downward gravitational force: [tex]\( F_{\text{gravity}} = \text{mass} \times g = 12.7 \, \text{kg} \times 9.81 \, \text{m/s}^2 = 124.587 \, \text{N} \)[/tex]
- Vertical component of the applied force: [tex]\( F_{\text{applied, vertical}} = 28.5 \, \text{N} \times \sin(60.3^\circ) = 24.7567 \, \text{N} \)[/tex]

The normal force (N) is reduced by the vertical component of the applied force:
[tex]\[ N = F_{\text{gravity}} - F_{\text{applied, vertical}} = 124.587 \, \text{N} - 24.7567 \, \text{N} = 99.83100183851155 \, \text{N} \][/tex]

### 4. Calculating the Force of Friction
The kinetic friction force (F_f) is found by multiplying the normal force (N) by the coefficient of kinetic friction (µₖ):
[tex]\[ \mu_k = 0.0802 \][/tex]
[tex]\[ F_f = \mu_k \times N = 0.0802 \times 99.83100183851155 \, \text{N} = 8.006446347448627 \, \text{N} \][/tex]

### Final Answer
Thus, the force of friction acting on the package is:
[tex]\[ F_f = 8.01 \, \text{N} \][/tex]

(Note: For clarity in reporting the final answer, values are often rounded to two decimal places, which is typical in many scientific contexts.)