Problem

Consider the ladder in Example 8.5 and assume there is friction between the vertical wall...

Consider the ladder in Example 8.5 and assume there is friction between the vertical wall and the ladder, with µs = 0.30. Find the angle ϕ at which the ladder just begins to slip.

Example 8.5

Leaning on a Wall

We continue our theme of housepainters and physics from Example 8.4 and consider the stability of a ladder of mass m and length L leaning against a wall as shown in Figure 8.23. The painter wants to be sure it will not slip across the floor. He knows that the frictional force exerted by the floor on the bottom end of the ladder is needed to keep the ladder in equilibrium. He therefore performs a careful experiment and finds that the coefficient of static friction between the ladder and the floor is μs = 0.50. He now wants to calculate the angle ϕ at which the ladder will just become unstable and slide across the floor. For simplicity, he assumes there is no frictional force between the ladder and the wall.

Figure 8.23 Example 8.5. Forces acting a leaning ladder. We assume there is no friction between the wall and the top end of the ladder.

RECOGNIZE THE PRINCIPLE

The ladder is in equilibrium, so we can apply the general conditions for static equilibrium The angle ϕ plays a key role: if we take the bottom of the ladder (point P) as our pivot point, the torque on the ladder due to gravity increases as ϕ becomes smaller because the lever arm increases. At some critical value of the angle, the torque will be large enough that the ladder will start to move.

SKETCH THE PROBLEM

Figure 8.23 shows all the forces acting on the ladder along with their components along x and y. There are normal forces exerted by the floor (NF) and the wall (Nw), the force of friction exerted by the floor (Ffriction), and the ladder’s weight (mg). The force from the floor acts on the end of the ladder at point P, while the force of gravity acts at the ladder’s center of mass.

IDENTIFY THE RELATIONSHIPS

Let’s choose the base of the ladder at P as the pivot point. By doing so, we make the torques due to the forces from the floor (NF and Ffriction) zero because these forces act at P and hence their lever arms are zero. Using the lever arms and angles from Figure 8.23, the torque condition is

where we have used the general definition of torque (Eq. 8.15) along with the trigonometric identity sin(90° – ϕ) = cos(ϕ). The appearance of the cosine term in Equation (1) can be understood from the expression for torque in Equation 8.16. The torque due to the ladder’s weight is the product of L/2 (the distance from the center of mass to the pivot point) and the component of the force of gravity perpendicular to the ladder. Figure 8.23 shows that this component involves cos ϕ.

Since the ladder is in static equilibrium, the sum of all forces along both x and y must be zero, which leads to

SOLVE

From Equation (3), we can immediately solve for the normal force from the floor and get NF = mg. We also know that the maximum frictional force is Ffriction = μsNF, so we have

Ffriction μsNF = μsmg

and, according to Equation (2), this result is also equal to Nw. Inserting into Equation (1) leads to

Using the value μs = 0.50 measured by the painter gives

What does it mean?

For a smaller angle, the frictional force would not be able to keep the ladder in equilibrium and the ladder would slip. Most sensible painters would probably choose a steeper angle than 45°.

Example 8.4

Painters and Physics

A housepainter is using a board of length L = 3.0 m that sits on two supports as a scaffolding on one side of a house (Fig. 8.22). The painter often wants to reach very far to one side of the board so that he can paint as much area as possible without moving the scaffolding, (a) How far can the painter in Figure 8.22 walk to the right-hand side of the board before it tips? The mass of the painter is 80 kg, and the mass of the board is 30 kg. (b) What is the force of the right-hand support on the board when the board just begins to tip? Assume the painter is always standing straight up, so his center of mass is always above his feet.

RECOGNIZE THE PRINCIPLE

The board does not quite tip, so it is always in rotational equilibrium. The total torque must therefore equal zero. The board is also in translational equilibrium, so the total force on the board must also equal zero.

SKETCH THE PROBLEM

Figure 8.22 shows all the forces on the board: the forces exerted by both supports FR and FL, the force of gravity on the board FB, and the force of gravity on the painter FP The figure also shows where each force acts, information that is needed to calculate the associated torques.

IDENTIFY THE RELATIONSHIPS

We choose the pivot point to be at P because that is the point around which the board would rotate if the painter moves too far to the right. When the painter is standing as far to the right as possible, the board will just barely begin to tip, which means that FL, the force exerted by the left-hand support, is zero.

We next compute the torques using the information in Figure 8.22 and apply the condition for rotational equilibrium ∑τ = 0. Using the lever arms from Figure 8.22 and FB = —mBg (the weight of the board) and FP = −mPg (the weight of the painter), we have

The signs of these torque terms are determined using the convention that a counterclockwise rotation is in the positive direction. For example, the last term on the right in Equation (1) is the torque from the weight of the painter; this torque would produce a clockwise rotation and thus is negative.

The next step is to write the condition for translational equilibrium . The forces on the board are all along the y direction. The force condition is then

Notice that some of these terms, such as FB = −mBg and FP = −mPg, are themselves negative. Inserting FL = 0 and our values of FB and FP into Equation (2) gives

SOLVE

(a) We can now solve Equation (1) to find x, the location of the painter. Inserting FL = 0 (because the board is on the verge of tipping), we find
Inserting the given values of the masses of the board and painter, along with L, gives
The painter would thus be well-advised to remain at a distance somewhat less than 0.28 m from the pivot point.
(b) To find the force from the right-hand support, we rearrange Equation (2) and get

What does it mean?

Since FL is zero, the right-hand support must support the entire weight of the system. Hence, the force from the right-hand support must equal the weight of the board plus the weight of the painter, as found in the answer to part (b).