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To solve this problem, we will break it down into steps. The waste container is in equilibrium, meaning the forces acting on it are balanced. Here's how we approach the problem:

### Step 1: Identify the forces involved
There are several forces acting on the waste container:
- **Weight (\( W \))**: This is the gravitational force acting vertically downwards. The weight is calculated using \( W = mg \), where \( m = 400 \, \text{kg} \) and \( g = 9.8 \, \text{m/s}^2 \). Thus, \( W = 400 \times 9.8 = 3920 \, \text{N} \).
- **Normal force (\( N \))**: This is the force exerted by the slope perpendicular to its surface.
- **Frictional force (\( F_f \))**: This resists motion along the slope and is given by \( F_f = \mu N \), where \( \mu = 0.3 \) is the coefficient of friction.
- **Tension (\( T \))**: This is the force exerted by the winch, acting up the slope to keep the container in equilibrium.

### Step 2: Break down the weight into components
The weight can be broken into two components:
- **Parallel to the slope**: \( W_{\parallel} = W \sin(\theta) = 3920 \times \sin(18^\circ) \)
- **Perpendicular to the slope**: \( W_{\perp} = W \cos(\theta) = 3920 \times \cos(18^\circ) \)

Let’s calculate these components first.

### Step 3: Calculate the weight components
We'll calculate \( W_{\parallel} \) and \( W_{\perp} \) for \( \theta = 18^\circ \).

\[
W_{\parallel} = 3920 \times \sin(18^\circ)
\]
\[
W_{\perp} = 3920 \times \cos(18^\circ)
\]

I'll compute these now.The components of the weight are:
- **Parallel to the slope**: \( W_{\parallel} = 1211.35 \, \text{N} \)
- **Perpendicular to the slope**: \( W_{\perp} = 3728.14 \, \text{N} \)

### Step 4: Calculate the frictional force
The frictional force (\( F_f \)) is given by:
\[
F_f = \mu N = \mu W_{\perp}
\]
Substituting the values:
\[
F_f = 0.3 \times 3728.14
\]

Let’s compute this now.The frictional force is \( F_f = 1118.44 \, \text{N} \).

### Step 5: Set up equilibrium conditions
Since the container is in equilibrium, the sum of forces along the slope must be zero. The tension force \( T \) and the frictional force \( F_f \) balance the component of the weight pulling the container down the slope \( W_{\parallel} \).

The net force condition is:
\[
T + F_f = W_{\parallel}
\]

The tension \( T \) must satisfy the inequality:
\[
W_{\parallel} - F_f \leq T \leq W_{\parallel} + F_f
\]

Substitute the values for \( W_{\parallel} \) and \( F_f \):
\[
1211.35 - 1118.44 \leq T \leq 1211.35 + 1118.44
\]

Let's compute the range of \( T \).The tension \( T \) must lie within the following range:
\[
92.90 \, \text{N} \leq T \leq 2329.79 \, \text{N}
\]

This means that the winch's tension should be at least 92.9 N to balance the forces, but it can go up to 2329.79 N.

Would you like further details or have any questions?

Here are five related questions to expand your understanding:
1. How would increasing the angle of the slope affect the tension required?
2. What happens to the range of tension if the coefficient of friction is increased?
3. How would the situation change if the mass of the container increased?
4. What if the slope was frictionless—how would you calculate the tension then?
5. How does the direction of the frictional force change if the container starts moving up the slope?

**Tip:** Always resolve forces into components when dealing with inclined planes to simplify calculations.

A waste container of mass 400kg is in equilibrium on a rough slope at an angle of 18° to the horizontal. The coefficient of friction between the slope and the skip is 0.3. It is held in equilibrium by a winch with tension T N. Find the range of possible values for T.

Learn how to calculate the range of possible tension values for a waste container on a rough slope with friction. This detailed step-by-step solution covers key physics concepts such as equilibrium, forces on inclined planes, and friction. Suitable for students in Grades 10-12.

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