A student tries to push a 27.6 kg desk across the room. She pushes at a downward angle of 36° (measured from the horizontal) with a force of 213 N. As a result, the desk moves with a constant velocity of 1.1 m/s.

How much friction is present? 

Ff = 
 N

How much does the desk weigh?

Fg = 
 N

How much Normal force must be present?

FN = 
 N

Let's break down the problem and solve for each of the requested values step by step:

### 1. **Frictional Force (Ff)**:

Since the desk is moving at a **constant velocity**, the net force acting on it must be zero (Newton's First Law). This implies that the force of friction **Ff** must exactly balance the horizontal component of the pushing force.

The horizontal component of the applied force \( F_{\text{push}} \) is:
\[
F_{\text{horizontal}} = F_{\text{push}} \cdot \cos(\theta)
\]
Where:
- \( F_{\text{push}} = 213 \, \text{N} \) (the applied force)
- \( \theta = 36^\circ \) (the angle with respect to the horizontal)

Thus, the frictional force \( F_f \) is:
\[
F_f = F_{\text{horizontal}} = 213 \cdot \cos(36^\circ)
\]

\[
F_f \approx 213 \cdot 0.8090 = 172.32 \, \text{N}
\]

### **Frictional Force** \( F_f = 172.32 \, \text{N} \)

---

### 2. **Weight of the Desk (Fg)**:

The weight of the desk is given by the force due to gravity acting on its mass. This is calculated as:
\[
F_g = m \cdot g
\]
Where:
- \( m = 27.6 \, \text{kg} \) (mass of the desk)
- \( g = 9.8 \, \text{m/s}^2 \) (acceleration due to gravity)

\[
F_g = 27.6 \cdot 9.8 = 270.48 \, \text{N}
\]

### **Weight of the Desk** \( F_g = 270.48 \, \text{N} \)

---

### 3. **Normal Force (FN)**:

The normal force is the perpendicular force exerted by the surface on the desk. The applied force has a vertical component that adds to the desk's weight, so the normal force must counteract both the desk’s weight and the downward vertical component of the applied force.

The vertical component of the applied force is:
\[
F_{\text{vertical}} = F_{\text{push}} \cdot \sin(\theta)
\]
\[
F_{\text{vertical}} = 213 \cdot \sin(36^\circ)
\]
\[
F_{\text{vertical}} \approx 213 \cdot 0.5878 = 125.19 \, \text{N}
\]

The normal force \( F_N \) is then the sum of the desk's weight and this vertical force:
\[
F_N = F_g + F_{\text{vertical}}
\]
\[
F_N = 270.48 + 125.19 = 395.67 \, \text{N}
\]

### **Normal Force** \( F_N = 395.67 \, \text{N} \)

---

### Summary:
- **Frictional Force (Ff):** \( 172.32 \, \text{N} \)
- **Weight of the Desk (Fg):** \( 270.48 \, \text{N} \)
- **Normal Force (FN):** \( 395.67 \, \text{N} \)

Would you like further details on any of these steps? 

Here are five related questions:
1. How would the frictional force change if the angle of the applied force were increased?
2. What would happen to the normal force if the student applied the same force upwards at an angle?
3. How is friction affected by the velocity of the desk?
4. Can you calculate the coefficient of friction in this scenario?
5. How would reducing the mass of the desk affect the forces involved?

**Tip:** Whenever there is constant velocity, the forces in the system are balanced, meaning the net force equals zero.

A student tries to push a 27.6 kg desk across the room. She pushes at a downward angle of 36° (measured from the horizontal) with a force of 213 N. As a result, the desk moves with a constant velocity of 1.1 m/s. How much friction is present? How much does the desk weigh? How much Normal force must be present?

Learn how to calculate the frictional force, weight, and normal force for a desk with a mass of 27.6 kg pushed at a 36° downward angle with a force of 213 N. The desk moves with a constant velocity, meaning forces are balanced. This problem involves Newton's Laws of Motion and trigonometry, suitable for high school students (Grades 9-12).

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