2. A worker pushes a 45.0-kg crate across a rough factory floor. The crate is very short, so the worker has to bend down to push on it, causing her to push down at an angle of 22.5° from horizontal. If the coefficient of static friction between the crate and the floor is 0.455, how big of a force is needed to start the crate moving?
2. A worker pushes a 45.0-kg crate across a rough factory floor. The crate is very short, so the worker has to bend down to push on it, causing her to push down at an angle of 22.5° from horizontal. If the coefficient of static friction between the crate and the floor is 0.455, how big of a force is needed to start the crate moving?
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ISBN:9781305952300
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Publisher:Raymond A. Serway, Chris Vuille
Chapter1: Units, Trigonometry. And Vectors
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Problem 1CQ: Estimate the order of magnitude of the length, in meters, of each of the following; (a) a mouse, (b)...
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![**Problem 2: Calculating the Force Needed to Start Moving a Crate**
A worker attempts to push a 45.0-kg crate across a rough factory floor. The crate's short height forces the worker to adopt a bent posture, resulting in a downward push at an angle of 22.5° from the horizontal plane. The coefficient of static friction between the floor and the crate is measured at 0.455. Determine the magnitude of the force required to initiate movement of the crate.
### Analysis:
To solve this problem, consider the forces acting on the crate:
1. **Weight**: The gravitational force acting downwards on the crate, calculated as mass (m) multiplied by gravitational acceleration (g = 9.81 m/s²).
2. **Normal Force**: The perpendicular force exerted by the floor on the crate.
3. **Applied Force**: The force applied by the worker, which has both horizontal and vertical components due to the angle.
4. **Static Frictional Force**: The resistive force preventing the crate from moving, calculated using the coefficient of static friction (μ_s) and the normal force.
### Approach:
1. **Calculate the Weight**:
\[
F_{\text{gravity}} = m \times g = 45.0 \, \text{kg} \times 9.81 \, \text{m/s}^2
\]
2. **Resolve the Applied Force into Components**:
- Horizontal component (\(F_{\text{x}}\)): \(F \cos(22.5^\circ)\)
- Vertical component (\(F_{\text{y}}\)): \(F \sin(22.5^\circ)\)
3. **Determine the Normal Force**:
- Normal force is adjusted by the vertical component of the applied force:
\[
N = F_{\text{gravity}} - F_{\text{y}}
\]
4. **Calculating Static Friction**:
- Static friction \(F_{\text{friction}}\) is given by:
\[
F_{\text{friction}} = \mu_s \times N
\]
5. **Setting Up the Equation**:
- To start the movement, the horizontal component of the applied force must overcome the static friction. Therefore:
\[
F_{\text{x}} = F_{\text{friction}}
\](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F278fbf48-8175-4c44-ac0e-2265809c567f%2F3bdbda67-ee7d-4659-aefb-2b823d42bde1%2Fjs5s3r_processed.png&w=3840&q=75)
Transcribed Image Text:**Problem 2: Calculating the Force Needed to Start Moving a Crate**
A worker attempts to push a 45.0-kg crate across a rough factory floor. The crate's short height forces the worker to adopt a bent posture, resulting in a downward push at an angle of 22.5° from the horizontal plane. The coefficient of static friction between the floor and the crate is measured at 0.455. Determine the magnitude of the force required to initiate movement of the crate.
### Analysis:
To solve this problem, consider the forces acting on the crate:
1. **Weight**: The gravitational force acting downwards on the crate, calculated as mass (m) multiplied by gravitational acceleration (g = 9.81 m/s²).
2. **Normal Force**: The perpendicular force exerted by the floor on the crate.
3. **Applied Force**: The force applied by the worker, which has both horizontal and vertical components due to the angle.
4. **Static Frictional Force**: The resistive force preventing the crate from moving, calculated using the coefficient of static friction (μ_s) and the normal force.
### Approach:
1. **Calculate the Weight**:
\[
F_{\text{gravity}} = m \times g = 45.0 \, \text{kg} \times 9.81 \, \text{m/s}^2
\]
2. **Resolve the Applied Force into Components**:
- Horizontal component (\(F_{\text{x}}\)): \(F \cos(22.5^\circ)\)
- Vertical component (\(F_{\text{y}}\)): \(F \sin(22.5^\circ)\)
3. **Determine the Normal Force**:
- Normal force is adjusted by the vertical component of the applied force:
\[
N = F_{\text{gravity}} - F_{\text{y}}
\]
4. **Calculating Static Friction**:
- Static friction \(F_{\text{friction}}\) is given by:
\[
F_{\text{friction}} = \mu_s \times N
\]
5. **Setting Up the Equation**:
- To start the movement, the horizontal component of the applied force must overcome the static friction. Therefore:
\[
F_{\text{x}} = F_{\text{friction}}
\
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