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.

Question

(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

Answer 1

In order to do part b, the first thing you have to do is consider the horizontal and vertical forces.

For the vertical forces (taking upward as positive) you have the normal reaction at the ground (we'll call this "Ng") which is acting up, and we have the weight of the ladder (750) acting down. These are the only vertical forces.

So we have Ng - 750 = 0 (as it is in equilibrium, the vertical forces must sum to 0)
So Ng = 750 We've now solved one of the three forces you need to label

The second force we can label, the frictional force, actually follows from the value we've just worked out. Recall that friction = coefficient of friction (u/mu) multiplied by the normal reaction (in this case, Ng, or the normal reaction on the floor). mu is given as 0.40 in the question.

So Friction = 0.40 * Ng
= 0.40 * 750
= 300N

Finally, to get the normal reaction at the wall, we consider the horizontal forces. We'll take forces acting the right as positive. We have the normal reaction of the ladder on the wall (we'll call "Nw") acting to the right, and we have the frictional force at the bottom of the ladder acting to the left. From this we get:

Nw - Friction = 0 (in equilibrium, sum to 0)
So, Nw = Friction
But we know the friction is 300N, so Nw = 300N

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For part C, begin by taking moments about the point on the ladder that is touching the wall

Clockwise moments: 750*sin(theta)*10 + 300*cos(theta)*20
Anticlockwise: 750*sin(theta)*20

In equilibrium the clockwise moments = anticlockwise moments

So, 50*sin(theta)*10 + 300*cos(theta)*20 = 750*sin(theta)*20

multiplying out the values we get:

7500sin(theta) + 6000cos(theta) = 1500sin(theta)
7500sin(theta) = 6000cos(theta)

Note that tan(theta) = sin(theta)/cos(theta)

so diving both sides by cos(theta) and then diving through by 7500, we get:

tan(theta) = 6000/7500
solving for theta by taking the tan inverse gives us 38.7