A 5 m long pole of 3 kg mass is placed against a smooth vertical wall. Under equilibrium condition, if the pole makes an angle of 37 degrees with the horizontal, what is the fricti... A 5 m long pole of 3 kg mass is placed against a smooth vertical wall. Under equilibrium condition, if the pole makes an angle of 37 degrees with the horizontal, what is the frictional force between the pole and horizontal surface?
Understand the Problem
The question is asking us to determine the frictional force acting on a pole that is placed against a smooth vertical wall while considering equilibrium conditions. We will analyze the forces and torques involved in the scenario, particularly focusing on the angle and the weight of the pole.
Answer
The frictional force is given by $F_f = \frac{W \cdot \cos(\theta)}{2 \cdot \sin(\theta)}$.
Answer for screen readers
The frictional force acting on the pole is given by:
$$ F_f = \frac{W \cdot \cos(\theta)}{2 \cdot \sin(\theta)} $$
Steps to Solve
- Identify Forces Acting on the Pole
The pole experiences three primary forces:
- The weight of the pole ($W$), acting downwards at its center of gravity.
- The normal force ($N_{wall}$) from the wall, acting horizontally on the pole.
- The frictional force ($F_f$) at the base, acting horizontally opposing the normal force from the wall.
- Draw a Free Body Diagram
Sketch a diagram representing the pole. Label the weight ($W$), the normal force ($N_{wall}$), and the frictional force ($F_f$). Show that the pole is in equilibrium, meaning that the sum of forces and the sum of torques must equal zero.
- Set Up the Equilibrium Equations
For vertical equilibrium:
$$ N_{ground} = W $$
Since the pole is resting on the ground, the normal force from the ground ($N_{ground}$) must balance the weight of the pole.
For horizontal equilibrium:
$$ F_f = N_{wall} $$
The frictional force must equal the normal force from the wall since there is no movement horizontally.
- Incorporate Torque for Further Analysis
Choose a pivot point at the base of the pole. The torque due to the weight ($W$) and the torque due to the normal force from the wall ($N_{wall}$) must balance each other for rotational equilibrium.
Calculating torques:
$$ \text{Torque}{weight} = W \cdot \frac{L}{2} \cdot \cos(\theta) $$
$$ \text{Torque}{normal} = N_{wall} \cdot L \cdot \sin(\theta) $$
Where:
- $L$ is the length of the pole.
- $\theta$ is the angle between the pole and the ground.
Set the torques equal for equilibrium:
$$ W \cdot \frac{L}{2} \cdot \cos(\theta) = N_{wall} \cdot L \cdot \sin(\theta) $$
- Solve for the Frictional Force
Rearranging the equation from step 4, we get:
$$ N_{wall} = \frac{W \cdot \cos(\theta)}{2 \cdot \sin(\theta)} $$
Now substitute $N_{wall}$ into the friction equilibrium equation:
$$ F_f = \frac{W \cdot \cos(\theta)}{2 \cdot \sin(\theta)} $$
- Calculate the Final Values
Depending on the known values of $W$ and $\theta$, you can now compute $F_f$. For simplicity, plug in numbers as needed.
The frictional force acting on the pole is given by:
$$ F_f = \frac{W \cdot \cos(\theta)}{2 \cdot \sin(\theta)} $$
More Information
The frictional force ensures the pole does not slip at the base while it is in contact with the vertical wall, showcasing the principles of static equilibrium. Understanding these forces is key in static mechanics applications.
Tips
- Forgetting to consider all forces acting on the pole, particularly missing the normal force from the ground.
- Not setting the equations based on the equilibrium conditions; it's crucial to balance forces and torques properly.
- Failing to draw a free body diagram, which can lead to confusion about the directions and magnitudes of forces.
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