A 50-kg sack of rice is pushed with uniform speed across a horizon- tal floor by a force directed 30° below the horizontal. The coeffi- cient of friction is 0.25. Find the magnitude of the force

To find the magnitude of the force \( F \) required to push a 50-kg sack of rice with uniform speed across a horizontal floor, we can break this problem down into two main components: the forces acting on the vertical and horizontal directions.

### Step-by-step Solution:

1. **Given Data:**
   - Mass of the sack \( m = 50 \, \text{kg} \)
   - Coefficient of friction \( \mu = 0.25 \)
   - Angle \( \theta = 30^\circ \)
   - Acceleration due to gravity \( g = 9.8 \, \text{m/s}^2 \)

2. **Forces in the vertical direction:**

   In the vertical direction, we have:
   - The normal force \( N \) exerted by the ground on the sack.
   - The gravitational force \( mg \) acting downward.
   - The vertical component of the applied force \( F \sin \theta \), which also acts downward.

   Since the sack moves horizontally with uniform speed, the vertical forces are balanced. Therefore, the net vertical force is zero:
   \[
   N = mg + F \sin \theta
   \]
   
3. **Forces in the horizontal direction:**

   In the horizontal direction, we have:
   - The horizontal component of the applied force \( F \cos \theta \).
   - The frictional force, which opposes motion and is given by \( f_{\text{friction}} = \mu N \).

   Since the sack moves with uniform speed, the horizontal forces are also balanced. Therefore, the net horizontal force is zero:
   \[
   F \cos \theta = f_{\text{friction}} = \mu N
   \]

4. **Solving for the normal force \( N \):**

   From the vertical balance equation, we have:
   \[
   N = mg + F \sin \theta
   \]
   Substituting this into the horizontal balance equation:
   \[
   F \cos \theta = \mu (mg + F \sin \theta)
   \]

5. **Solving for \( F \):**

   Now, solve for \( F \):
   \[
   F \cos \theta = \mu mg + \mu F \sin \theta
   \]
   Rearranging the equation to isolate \( F \):
   \[
   F (\cos \theta - \mu \sin \theta) = \mu mg
   \]
   \[
   F = \frac{\mu mg}{\cos \theta - \mu \sin \theta}
   \]

6. **Substitute the values:**
   \[
   F = \frac{(0.25)(50)(9.8)}{\cos 30^\circ - (0.25) \sin 30^\circ}
   \]
   First, calculate the trigonometric values:
   \[
   \cos 30^\circ \approx 0.866, \quad \sin 30^\circ = 0.5
   \]
   Now substitute these into the equation:
   \[
   F = \frac{(0.25)(50)(9.8)}{0.866 - (0.25)(0.5)}
   \]
   \[
   F = \frac{122.5}{0.866 - 0.125} = \frac{122.5}{0.741}
   \]
   \[
   F \approx 165.3 \, \text{N}
   \]

### Final Answer:
The magnitude of the force \( F \) is approximately **165.3 N**.

Would you like any further clarification or additional details?

### 5 Related Questions:
1. How does the angle of applied force affect the force required to move the object?
2. What happens if the coefficient of friction increases in this problem?
3. How would the solution change if the sack were pushed at an angle above the horizontal?
4. Can you explain how to calculate the normal force in other friction problems?
5. How would you approach the problem if the sack were accelerating instead of moving at uniform speed?

### Tip:
Always resolve forces into their horizontal and vertical components when dealing with forces at an angle.

A 50-kg sack of rice is pushed with uniform speed across a horizontal floor by a force directed 30° below the horizontal. The coefficient of friction is 0.25. Find the magnitude of the force.

Learn how to calculate the force required to push a 50-kg sack of rice with uniform speed across a horizontal floor. This step-by-step solution uses friction and trigonometry concepts and applies Newton's laws of motion. Suitable for high school students in Grades 10-12.

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