QuestionJune 16, 2025

A 14.5 kg barrel is pulled with a 22.1 N force at a 48.7^circ angle,, across the ground where the coefficient of friction (mu _(k)) is equal to 0.200. What is the normal force acting upon the barrel? n=[?]N

A 14.5 kg barrel is pulled with a 22.1 N force at a 48.7^circ angle,, across the ground where the coefficient of friction (mu _(k)) is equal to 0.200. What is the normal force acting upon the barrel? n=[?]N
A 14.5 kg barrel is pulled with a 22.1 N force at a 48.7^circ angle,, across the ground where the coefficient of friction (mu _(k)) is equal to 0.200. What is the normal force acting upon the barrel? n=[?]N

Solution
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Answer

n = 117.6 \, \text{N} - 16.6 \, \text{N} = 101.0 \, \text{N} Explanation 1. Calculate the vertical component of the pulling force The vertical component is F_{y} = F \cdot \sin(\theta). Here, F = 22.1 \, \text{N} and \theta = 48.7^\circ. So, F_{y} = 22.1 \cdot \sin(48.7^\circ). 2. Calculate gravitational force Gravitational force is F_g = m \cdot g, where m = 14.5 \, \text{kg} and g = 9.8 \, \text{m/s}^2. So, F_g = 14.5 \cdot 9.8. 3. Calculate normal force Normal force n = F_g - F_{y}. Substitute F_g and F_{y} from previous steps.

Explanation

1. Calculate the vertical component of the pulling force<br /> The vertical component is $F_{y} = F \cdot \sin(\theta)$. Here, $F = 22.1 \, \text{N}$ and $\theta = 48.7^\circ$. So, $F_{y} = 22.1 \cdot \sin(48.7^\circ)$.<br />2. Calculate gravitational force<br /> Gravitational force is $F_g = m \cdot g$, where $m = 14.5 \, \text{kg}$ and $g = 9.8 \, \text{m/s}^2$. So, $F_g = 14.5 \cdot 9.8$.<br />3. Calculate normal force<br /> Normal force $n = F_g - F_{y}$. Substitute $F_g$ and $F_{y}$ from previous steps.
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