Question

A ladder of length 2L and mass M is positioned on level ground leaning against a wall such

that the angle between the ladder and the horizontal is α. The coefficient of static friction

between the ladder and the wall and between the ladder and the ground is μstatic = 0.65. The

centre of mass of the ladder is halfway along it.

(b) Draw a diagram indicating all the forces acting on the ladder, and state what each force

represents. Also show the direction of the x- and y-axes you will use. (Hint: there are 2

forces acting at the top of the ladder, 2 forces acting on the bottom of the ladder and 1 force

acting at its centre.)

(c) For the ladder to be in mechanical equilibrium:

(i) Write down equations for the total x- and y-components of the 5 forces acting on the

ladder.

(ii) Consider torques about the centre of the ladder. In which direction (into or out of the

page) does the torque due to each of the 5 forces force act?

(iii) Write down an equation for the sum of the torques about the centre of mass of the

ladder.

(iv) Use your equation for the torques to derive an expression for tan α in terms of the

magnitudes of the forces acting.

(d)(i) If the ladder is just on the point of slipping at both the upper and lower ends, what

can you say about the pair of forces acting at each of the top and bottom of the ladder?

(ii) Hence use this information, with the information from (c)(i) and the expression you have

derived in (c)(iv) to calculate the minimum angle that the ladder can form with the ground

in order for it not to slip. (Hint : In the expression for tan α, you will need to write each of

the forces in terms of one of the normal reaction forces, which then cancel out, leaving an

expression for tan α in terms of μstatic only.)

Answer 1

  α= A , μ = u

C)

ii) the ladder has tendency to slip down the wall so the torque required to slip the ladder down the wall must act into the plane.

iii) considering into the plane as positive direction

When ladder is slipping down

Torque(N1) = -(N1)(L) (SINA)

Torque(N2) = (N2)(L )(COSA)

Torque(f1) = -(F1)(L)(COSA)

Torque(f2) = -(F2)(L)(SINA)

Torque(mg) = 0

By adding all this torques you will get

L[ COSA(N2 - f1) - SINA(N1 + f2) ] ......(1)

iv)

Net torque = 0 for rotational equilibrium

Simplyfing equation 1 we get

TANA =( N2-f1)/ (N1 +f2)

D)

i)

For tranlational equilibrium in both directions

Horizontal f2= N1

Vertical N2 +f1 = mg

ii) using results from

D (i) & C (iv)

we get

TANA = (N2 - uN1)/ (N1+ uN2)

Now putting N1= uN2 and cancelling N2 from numerator and denominator we will get

TANA = (1-u²)/2u

When you put value of u in equation we get

A = 23.95 this is the minimum angle the ladder can make with horizontal