A uniform ladder of length 20.0m and weight 750 N is propped up against a smooth vertical wall with its lower end on a rough horizontal surface. The coefficient of friction between the ladder and this horizontal surface is 0.40.

(a) Draw a diagram for the ladder showing all the forces acting on the ladder.

(b) Work out and add the numerical values of each force clearly showing your justification in each case.

(c) Hence, calculate a value for the angle between the ladder and the wall if the ladder just remains in stable equilibrium.

a. can't do drawings here,

On the wall, there is a horizontal force out of the wall at the top of the ladder
On the base, there is a vertical force from the floor upwards, and a horizontal force (friction) pushing inward towards the wall
At ladder midength, there is a force dowward(weight).
that is your diagram.

So now sum vertical forces: You only have two (weightladder and floor upward).
Sum horizontal foces: wall force and friction at base.
Sum moments about any point: I recommend the floor contact point, you have ladder weight, and wall force then.

The angle should come put pretty quickly on the moment equation.

I have completed a but I just don't understand b and c. Please help me asap!!!!

(a) Here is a diagram illustrating the forces acting on the ladder:

<--- Wall
_________
| |
| |
| ↓ |
|Ladder |
| |
| |
|_________|

In this diagram, the ladder is represented as a vertical line, with the wall on the left side (indicated by the arrow) and the rough horizontal surface at the bottom.

The forces acting on the ladder are:
1. Weight (W): This force acts vertically downwards from the center of gravity of the ladder, which we can assume to be located at its midpoint.
2. Normal Reaction (N): This force acts perpendicular to the rough horizontal surface and supports the weight of the ladder.
3. Frictional Force (f): This force opposes the motion or tendency of motion between the ladder and the rough horizontal surface.
4. Applied Force (A): This force represents any external force applied to the ladder, such as someone pushing or pulling it.

(b) Let's work out the numerical values of each force:

1. Weight (W): Given that the weight of the ladder is 750 N, this force can be assumed to act straight downwards from the center of gravity. So, W = 750 N.

2. Normal Reaction (N): The normal reaction will be equal in magnitude and opposite in direction to the component of the weight that is perpendicular to the rough horizontal surface. Using trigonometry, we can find that N = W * cos(θ), where θ is the angle between the ladder and the horizontal surface. Since the ladder is in stable equilibrium, we can assume that the angle is less than 90 degrees, so the cosine of the angle is positive. Therefore, N = W * cos(θ) = 750 * cos(θ).

3. Frictional Force (f): The frictional force is given by the equation f = μN, where μ is the coefficient of friction between the ladder and the horizontal surface. Given that the coefficient of friction is 0.40, we can calculate f = 0.40 * N = 0.40 * (750 * cos(θ)).

4. Applied Force (A): If there is no external force being applied to the ladder, then A = 0.

(c) To calculate the angle between the ladder and the wall for stable equilibrium, we need to consider the sum of the moments acting on the ladder. The net moment must be zero for the ladder to be in equilibrium. In this case, the only moment acting on the ladder is due to the weight, W.

The moment arm of the weight is the distance from the point of contact with the horizontal surface to the center of gravity of the ladder, which is half its length. So, the moment arm is 20.0 m / 2 = 10.0 m.

The weight creates a clockwise moment, while the normal reaction and frictional force create counterclockwise moments. The equation for the sum of the moments is:

Clockwise moment = Counterclockwise moment
(W)(moment arm) = (N)(moment arm) + (f)(moment arm)

Substituting the values we have:
(750 N)(10.0 m) = (750 * cos(θ))(10.0 m) + (0.40 * (750 * cos(θ)))(10.0 m)

Simplifying and rearranging the equation:
7500 N*m = 7500 N*cos(θ) + 300 * N*cos(θ)

Combining like terms:
0 = 7800 N*cos(θ)

To solve for the angle θ, we can divide both sides of the equation by 7800 N:
0 = cos(θ)

Since cos(θ) = 0 only when θ = 90 degrees, the ladder will be in stable equilibrium when it forms a right angle with the wall.