Question

A uniform ladder of length 10m and mass 16 kg resting against a vertical wall making an angle of 37 with it. The vertical wall is frictionless ladder a point 8m lower end minium coefficient

Answer 1

The minimum coefficient is 0.55

Explanation:

If the electrician works safely, the ladder + man system should be in equilibrium.

For equilibrium ΣF= 0 and Στ = 0

for ΣFx = 0,Na = f

for ΣFY = 0,NB = mg + Mg = 60 × 10 + 16 × 1 = 760 N

Taking torque about B:

NA10cos(37∘) = mg8sin(37∘) + mg5sin(37∘)

NA = tan(37∘) / 10 (65×10×8+16×10×5)

From i f=NA=420 N

For maximum value of coefficient of friction.

f ≤ μNB

μ ≥ f NB = 420 / 760 = 0.55

Thus the minimum coefficient is 0.55

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Answer 2

Complete question:

A uniform ladder of length 10.0 m and mass 16.0 kg is resting against a vertical wall making an angle of 37° with it. The vertical wall is frictionless but the ground is rough. An electrician weighing 60.0 kg climbs up the ladder. If he stays on the ladder at a point 8.00 m from the lower end, what will be the normal force and the force of friction on the ladder by the ground ? What should be the minimum coefficient of friction for the electrician to work safely ?

Answer:

The normal force and the force of friction on the ladder is 744.8 N and 412 N respectively.

The minimum coefficient of friction for the electrician is 0.553

Explanation:

The force F₁ and F₂ are acting on the ladder.

The ground is rough, thus there is no slip or rotation.

The torque is given by the expression:

mg (8 sin 37°) + Mg (5 sin 37°) - F₂ (10 cos 37°) = 0

(60 × 9.8 × 8 sin 37°) + (16 × 9.8 × 5 sin 37°) - F₂ (10 cos 37°) = 0

On calculating all the numbers, we get,

F₂ = 412 N

Since, there is no slip, then the force F₂ acting on the ladder is equal to frictional force.

∴ F₂ = 412 N = Frictional force

The normal force is given as:

F₁ = (m + M)g

F₁ = (60 + 16) × 9.8

∴ F₁ = 744.8 N

Now, the minimum coefficient of friction is given by the formula:

μ = F₂/F₁

μ = 412/744.8

∴ μ = 0.553