To solve the problem, we need to analyze the forces acting on the crate. Let's outline the steps to determine the normal force, friction force, and coefficient of friction.
1. Normal Force on the Box
The crate is moving down the wall under the influence of gravity and the person pushing against it horizontally. The forces acting on the crate in the vertical direction are:
- The gravitational force acting downward: \( F_{\text{gravity}} = mg \)
- The normal force acting from the wall, which we will denote as \( N \).
Calculating the gravitational force where \( g \approx 9.81 , \text{m/s}^2 \): \[ F_{\text{gravity}} = m \cdot g = 111 , \text{kg} \cdot 9.81 , \text{m/s}^2 = 1093.91 , \text{N} \]
Since the crate is moving vertically downward with a constant acceleration (6.0 m/s²), the net force in the vertical direction can be calculated using Newton's second law \( F_{net} = ma \):
\[ F_{net} = m \cdot a = 111 , \text{kg} \cdot 6.0 , \text{m/s}^2 = 666 , \text{N} \]
The net force is downward and can be expressed as: \[ F_{\text{gravity}} - N = F_{net} \]
Substituting the known values: \[ 1093.91 , \text{N} - N = 666 , \text{N} \] \[ N = 1093.91 , \text{N} - 666 , \text{N} \] \[ N \approx 427.91 , \text{N} \]
2. Friction Force Acting on the Box
The friction force \( F_{\text{friction}} \) can be determined from the horizontal force applied by the person and the normal force:
- The maximum static or kinetic friction force is given by \( F_{\text{friction}} = \mu N \) where \( \mu \) is the coefficient of friction. However, since we have a horizontal applied force, we can consider that \( F_{\text{friction}} \) is equal to the force applied by the person to stop the box.
Given that the person pushes horizontally with a force of \( F_{\text{push}} = 401 , \text{N} \), the friction force acting on the box is equal to this applied force, as there is no other horizontal force acting:
\[ F_{\text{friction}} = F_{\text{push}} = 401 , \text{N} \]
3. Coefficient of Friction Between the Wall and the Crate
The coefficient of friction \( \mu \) can be calculated by relating the friction force to the normal force:
\[ \mu = \frac{F_{\text{friction}}}{N} \]
Substituting the known values: \[ \mu = \frac{401 , \text{N}}{427.91 , \text{N}} \] \[ \mu \approx 0.935 \]
Summary of Findings
- Normal force on the box: \( \approx 427.91 , \text{N} \)
- Friction force acting on the box: \( 401 , \text{N} \)
- Coefficient of friction between the wall and the crate: \( \approx 0.935 \)