Question
A force of 1760 N is required to start moving a bundle of wooden planks up a ramp. If the ramp’s
incline is 17° and the mass of the planks is 266 kg, what is the coefficient of static friction between the
planks and the ramp?
incline is 17° and the mass of the planks is 266 kg, what is the coefficient of static friction between the
planks and the ramp?
Answers
GPT-4o mini
To find the coefficient of static friction (\( \mu_s \)) between the planks and the ramp, we can use the following relation that considers the forces acting on the planks on an incline:
\[
F_{\text{friction}} = \mu_s \cdot N
\]
where:
- \( F_{\text{friction}} \) is the force required to overcome static friction (1760 N in this case),
- \( N \) is the normal force.
The normal force (\( N \)) on an incline can be calculated using the following equation:
\[
N = mg \cos(\theta)
\]
where:
- \( m \) is the mass of the planks (266 kg),
- \( g \) is the acceleration due to gravity (approximately \( 9.81 \, \text{m/s}^2 \)),
- \( \theta \) is the angle of the incline (17°).
First, we calculate the normal force:
\[
N = 266 \, \text{kg} \times 9.81 \, \text{m/s}^2 \times \cos(17^\circ)
\]
Calculating \( \cos(17^\circ) \):
\[
\cos(17^\circ) \approx 0.9563
\]
Now substituting:
\[
N = 266 \times 9.81 \times 0.9563 \approx 266 \times 9.38 \approx 2495.68 \, \text{N}
\]
Now, substituting \( N \) into the equation for static friction:
\[
1760 \, \text{N} = \mu_s \cdot 2495.68 \, \text{N}
\]
Now, we can solve for \( \mu_s \):
\[
\mu_s = \frac{1760 \, \text{N}}{2495.68 \, \text{N}} \approx 0.706
\]
Thus, the coefficient of static friction between the planks and the ramp is approximately:
\[
\boxed{0.706}
\]
\[
F_{\text{friction}} = \mu_s \cdot N
\]
where:
- \( F_{\text{friction}} \) is the force required to overcome static friction (1760 N in this case),
- \( N \) is the normal force.
The normal force (\( N \)) on an incline can be calculated using the following equation:
\[
N = mg \cos(\theta)
\]
where:
- \( m \) is the mass of the planks (266 kg),
- \( g \) is the acceleration due to gravity (approximately \( 9.81 \, \text{m/s}^2 \)),
- \( \theta \) is the angle of the incline (17°).
First, we calculate the normal force:
\[
N = 266 \, \text{kg} \times 9.81 \, \text{m/s}^2 \times \cos(17^\circ)
\]
Calculating \( \cos(17^\circ) \):
\[
\cos(17^\circ) \approx 0.9563
\]
Now substituting:
\[
N = 266 \times 9.81 \times 0.9563 \approx 266 \times 9.38 \approx 2495.68 \, \text{N}
\]
Now, substituting \( N \) into the equation for static friction:
\[
1760 \, \text{N} = \mu_s \cdot 2495.68 \, \text{N}
\]
Now, we can solve for \( \mu_s \):
\[
\mu_s = \frac{1760 \, \text{N}}{2495.68 \, \text{N}} \approx 0.706
\]
Thus, the coefficient of static friction between the planks and the ramp is approximately:
\[
\boxed{0.706}
\]