Question
A 120 N force acts on a 100 kg mass at a 30-degree angle. If the coefficient of sliding friction is 0.1, how do I find the acceleration?
Answers
GPT-4o mini
To find the acceleration of the mass acted upon by a force at an angle, we need to calculate the net force acting on the mass and then apply Newton's second law.
1. **Calculate the gravitational force** (\( F_g \)):
\[
F_g = m \cdot g
\]
where
\( m = 100 \, \text{kg} \) (mass) and
\( g = 9.81 \, \text{m/s}^2 \) (acceleration due to gravity).
\[
F_g = 100 \, \text{kg} \cdot 9.81 \, \text{m/s}^2 = 981 \, \text{N}
\]
2. **Resolve the applied force** (\( F_a = 120 \, \text{N} \)) into its horizontal and vertical components:
- The horizontal component \( F_{ax} \):
\[
F_{ax} = F_a \cdot \cos(\theta) = 120 \, \text{N} \cdot \cos(30^\circ)
\]
\[
F_{ax} = 120 \, \text{N} \cdot \frac{\sqrt{3}}{2} \approx 103.92 \, \text{N}
\]
- The vertical component \( F_{ay} \):
\[
F_{ay} = F_a \cdot \sin(\theta) = 120 \, \text{N} \cdot \sin(30^\circ)
\]
\[
F_{ay} = 120 \, \text{N} \cdot 0.5 = 60 \, \text{N}
\]
3. **Calculate the normal force** (\( F_N \)):
The normal force is affected by the weight of the mass and the vertical component of the applied force:
\[
F_N = F_g - F_{ay} = 981 \, \text{N} - 60 \, \text{N} = 921 \, \text{N}
\]
4. **Calculate the frictional force** (\( F_f \)):
The frictional force can be calculated using the coefficient of friction \( \mu = 0.1 \):
\[
F_f = \mu \cdot F_N = 0.1 \cdot 921 \, \text{N} = 92.1 \, \text{N}
\]
5. **Calculate the net force** (\( F_{net} \)):
The net force acting on the mass in the horizontal direction is the horizontal component of the applied force minus the frictional force:
\[
F_{net} = F_{ax} - F_f
\]
\[
F_{net} = 103.92 \, \text{N} - 92.1 \, \text{N} = 11.82 \, \text{N}
\]
6. **Calculate the acceleration** (\( a \)):
Using Newton's second law \( F = m \cdot a \):
\[
a = \frac{F_{net}}{m} = \frac{11.82 \, \text{N}}{100 \, \text{kg}} = 0.1182 \, \text{m/s}^2
\]
Thus, the acceleration of the mass is approximately:
\[
\boxed{0.12 \, \text{m/s}^2}
\]
1. **Calculate the gravitational force** (\( F_g \)):
\[
F_g = m \cdot g
\]
where
\( m = 100 \, \text{kg} \) (mass) and
\( g = 9.81 \, \text{m/s}^2 \) (acceleration due to gravity).
\[
F_g = 100 \, \text{kg} \cdot 9.81 \, \text{m/s}^2 = 981 \, \text{N}
\]
2. **Resolve the applied force** (\( F_a = 120 \, \text{N} \)) into its horizontal and vertical components:
- The horizontal component \( F_{ax} \):
\[
F_{ax} = F_a \cdot \cos(\theta) = 120 \, \text{N} \cdot \cos(30^\circ)
\]
\[
F_{ax} = 120 \, \text{N} \cdot \frac{\sqrt{3}}{2} \approx 103.92 \, \text{N}
\]
- The vertical component \( F_{ay} \):
\[
F_{ay} = F_a \cdot \sin(\theta) = 120 \, \text{N} \cdot \sin(30^\circ)
\]
\[
F_{ay} = 120 \, \text{N} \cdot 0.5 = 60 \, \text{N}
\]
3. **Calculate the normal force** (\( F_N \)):
The normal force is affected by the weight of the mass and the vertical component of the applied force:
\[
F_N = F_g - F_{ay} = 981 \, \text{N} - 60 \, \text{N} = 921 \, \text{N}
\]
4. **Calculate the frictional force** (\( F_f \)):
The frictional force can be calculated using the coefficient of friction \( \mu = 0.1 \):
\[
F_f = \mu \cdot F_N = 0.1 \cdot 921 \, \text{N} = 92.1 \, \text{N}
\]
5. **Calculate the net force** (\( F_{net} \)):
The net force acting on the mass in the horizontal direction is the horizontal component of the applied force minus the frictional force:
\[
F_{net} = F_{ax} - F_f
\]
\[
F_{net} = 103.92 \, \text{N} - 92.1 \, \text{N} = 11.82 \, \text{N}
\]
6. **Calculate the acceleration** (\( a \)):
Using Newton's second law \( F = m \cdot a \):
\[
a = \frac{F_{net}}{m} = \frac{11.82 \, \text{N}}{100 \, \text{kg}} = 0.1182 \, \text{m/s}^2
\]
Thus, the acceleration of the mass is approximately:
\[
\boxed{0.12 \, \text{m/s}^2}
\]