A ladder (ℓL = 7.50 m) of weight WL = 350 N leans against a smooth vertical wall. The term "smooth" means that the wall can exert only a normal force directed perpendicular to the wall and cannot exert a frictional force parallel to it. A firefighter, whose weight is 886 N, stands 6.50 m up from the bottom of the ladder (this distance goes along the ladder, it is not the vertical height). Assume that the ladder's weight acts at the ladder's center, and neglect the hose's weight. What is the minimum value for the coefficient of static friction between the ladder and the ground, so that the ladder (with the fireman on it) does not slip? (Assume θ = 51.0°.)
Why did the ladder join the circus? Because it wanted to show off its balancing skills!
To solve this problem, we need to consider the balance of forces acting on the ladder. The ladder is in equilibrium, meaning that the net force acting on it is zero.
Let's take a look at the forces involved:
1. Weight of the ladder (WL): Since the weight is acting at the center of the ladder, we can consider it as a single force acting downwards from the center of the ladder. This force can be calculated using WL = m*g, where g is the acceleration due to gravity.
2. Normal force from the wall (NW): Since the wall can only exert a normal force perpendicular to it, this force counteracts the weight of the ladder and acts vertically upward.
3. Normal force from the ground (NG): This force is exerted by the ground on the ladder, counteracting the weight of the ladder and the firefighter. This force is perpendicular to the ground.
4. Frictional force from the ground (FG): This force acts parallel to the ground and opposes the tendency of the ladder to slip.
To find the minimum coefficient of static friction (μs), we need to consider the point at which slipping would start to occur. This happens when the frictional force reaches its maximum value, given by FG(max) = μs * NG.
To analyze the forces, we can break them down into their components:
1. Weight of the ladder:
WL(x) = -WL * sin(θ)
WL(y) = WL * cos(θ)
2. Normal force from the wall:
NW(x) = 0
NW(y) = -WL(y) = -WL * cos(θ)
3. Normal force from the ground:
NG(x) = 0
NG(y) = WL(y) + WF(y) - NG = WL * cos(θ) + WF - NG
4. Frictional force from the ground:
FG(x) = -μs * NG = -μs * (WL * cos(θ) + WF - NG)
FG(y) = 0
Since the ladder is in equilibrium, the sum of the forces in the x and y directions must be zero:
∑F(x) = 0: -μs * (WL * cos(θ) + WF - NG) = 0
∑F(y) = 0: -WL * cos(θ) + WL * cos(θ) + WF - NG = 0
Simplifying these equations, we get:
-μs * (WL * cos(θ) + WF - NG) = 0
WF - NG = 0
Substituting NG = WL * cos(θ) + WF into the first equation, we have:
-μs * (WL * cos(θ) + WF - (WL * cos(θ) + WF)) = 0
Simplifying further, we get:
0 = 0
As we can see, this equation holds true regardless of the value of μs. This means that the ladder will not slip as long as there is some friction present, regardless of the value of μs. Therefore, the minimum value for the coefficient of static friction μs is 0.
Remember, safety first, but let's hope the ladder doesn't try any acrobatics!
To solve this problem, we first need to understand the forces acting on the ladder.
1. Weight of the ladder (WL = 350 N): The weight of the ladder acts downward at its center.
2. Normal force exerted by the wall: The wall only exerts a normal force on the ladder, which is directed perpendicular to the wall.
3. Normal force exerted by the ground (N): The ground exerts a normal force acting perpendicular to the ground surface.
4. Frictional force between the ladder and the ground (f): The frictional force acts parallel to the ground surface and opposes the motion of the ladder.
Let's assume the coefficient of static friction between the ladder and the ground is μ.
Now, we can write the equations of motion along the ladder and perpendicular to the ladder.
Along the ladder:
∑F∥ = 0
N - WL - WF = 0 ---- (1)
Perpendicular to the ladder:
∑F⊥ = 0
N - NF = 0 ---- (2)
The ladder is in equilibrium, which means there is no net torque acting around any point. We can consider the torque about the bottom of the ladder.
∑τ(bottom) = 0
τL - τF = 0 ---- (3)
The torque due to the weight of the ladder (τL) is given by:
τL = WL * (ℓL/2) * sin(θ) ---- (4)
The torque due to the firefighter's weight (τF) is given by:
τF = WF * (ℓF) * sin(θ) ---- (5)
Substituting equations (4) and (5) into equation (3), we get:
WL * (ℓL/2) * sin(θ) - WF * (ℓF) * sin(θ) = 0
Simplifying the equation, we find:
ℓL * WL/2 - ℓF * WF = 0
ℓL * 350/2 - ℓF * 886 = 0
(7.50 * 350)/2 - 6.50 * 886 = 0
2625 - 5761 = 0
-3136 = 0
As you can see, there is a contradiction because the equation (-3136 = 0) is not true. This means that the frictional force is not sufficient to prevent the ladder from slipping, regardless of the value of the coefficient of static friction.
Therefore, the minimum value of the coefficient of static friction is not applicable, and the ladder will slip, no matter how high the coefficient of static friction is.
To find the minimum value for the coefficient of static friction between the ladder and the ground, we need to analyze the forces acting on the ladder.
First, let's draw a free-body diagram showing all the forces acting on the ladder:
```
|
|
| F_friction <-- Ground
|
|
F_normal <-- Ground |
o---------------------------L
|
| θ
|
| F_gravity <-- Ladder's weight
|
| F_firefighter <-- Firefighter's weight
|
```
1. F_gravity: The weight of the ladder (WL = 350 N) acts downwards and is located at the center of the ladder.
2. F_normal: The normal force exerted by the ground acts perpendicular to the wall. It counterbalances the vertical component of F_gravity and the vertical component of F_firefighter.
3. F_firefighter: The firefighter's weight (WF = 886 N) acts downwards and applies a force at a distance of 6.50 m up from the bottom of the ladder.
4. F_friction: The frictional force between the ladder and the ground opposes the horizontal component of F_gravity and the horizontal component of F_firefighter.
Now, let's resolve the forces vertically and horizontally:
Vertical Equilibrium: F_normal - F_gravity - F_firefighter = 0
Horizontal Equilibrium: F_friction - F_gravity*sin(θ) - F_firefighter*sin(θ) = 0
Using trigonometric identities:
F_gravity*sin(θ) = WL*sin(θ)
F_firefighter*sin(θ) = WF*sin(θ)
Substituting these values into the horizontal equilibrium equation:
F_friction - WL*sin(θ) - WF*sin(θ) = 0
Now we can solve for F_friction:
F_friction = WL*sin(θ) + WF*sin(θ)
Finally, we can determine the minimum value for the coefficient of static friction (μs) using the equation:
F_friction = μs * F_normal
Substituting the values we have:
μs * (F_gravity + F_firefighter) = WL*sin(θ) + WF*sin(θ)
Simplifying the equation and substituting the given values:
μs * (350 N + 886 N) = 350 N * sin(51.0°) + 886 N * sin(51.0°)
Now, we can solve for μs:
μs = (350 N * sin(51.0°) + 886 N * sin(51.0°)) / (350 N + 886 N)
Calculating the value, we get:
μs ≈ 0.702
Therefore, the minimum value for the coefficient of static friction between the ladder and the ground is approximately 0.702.
There are only two horizontal forces in this problem, at the wall and at the ground.
Therefore they are equal and opposite, call the magnitude F
now balance vertical forces
886 + 350 = force down = 1236 N
that is therefore the normal force up
at the ladder base
the max friction force is thus
1236 mu
now take moments about top of ladder
clockwise:
1236 * 7.5 cos 51
counterclockwise
1236 mu *7.5 sin 51 + 350 *3.75 cos 51 + 886 *1*cos 51
set those equal and solve for mu
by the way look at the old questions answered below