Answer to Question #163668 in Mechanics | Relativity for particle vs rigid bodies

Question

A uniform ladder rests against a smooth wall so that it makes an angle of 60⁰ with the ground. The ladder is 10.0 m long and weighs 150 N. How far can a 250 N child go up the ladder before it gets slips?

The coefficient of friction between the ladder and the ground is 0.4.

Answer 1

𝜇1 = 0.400 − coefficient of friction between the ladder and the ground;

𝜇2 = 0.450 − coefficient of friction between the ladder and the wall;

𝛼 = 60° − angle which ladder makes with the ground;

𝑁1 − reaction force from the ground;

𝑁2 − reaction force from the wall;

𝑃1 = 150𝑁 − weight of the ladder;

𝑃2 = 250𝑁 − weight of the child;

𝐿 = 10.0m − length of the ladder.

We will consider the extreme case when the person is standing at a maximum distance d from the beginning of the ladder.

Newton's second law for the ladder (the first law of equilibrium):

$\vec{F_{fr1}}+\vec{F_{fr2}}+\vec{P_1}+\vec{P_2}+\vec{N_1}+\vec{N_2}=\vec{0},$

Projection of the law on the X-axis:

$N_2-F_{fr1}=0,$ (1)

projection of the law on the X-axis:

$N_1+F_{fr2}-P_1-P_2=0.$ (2)

Law of dry friction:

$F_{fr1}=\mu_1N_1$ (3)

$F_{fr2}=\mu_2N_2$ (4)

(3) and (4) in (1) and (2):

(3) → (1):  $N_2-\mu_1N1=0$

$N_1=\frac{N_2}{\mu_1}$ (5)

(4) → (2): $N_1+\mu_2N_2-P_1-P_2=0$

(5) → (2):

$\frac{N_2}{\mu_1}+\mu_2N_2-P_1-P_2=0$ ,

$N_2=\frac{\mu_1(P_1+P_2)}{1+\mu_1\mu_2}=135.6~\text{N},$

$F_{fr2}=\mu_2N_2=61.02~\text{N}.$

The second law of equilibrium:

$M_{P_1}+M_{P_2}+M_{F_{fr2}}+M_{N_2}=0,$ (6)

($M_{N_1}=M_{F_{fr1}}=0$ 𝑏𝑒𝑐𝑎𝑢𝑠𝑒 𝑚𝑜𝑚𝑒𝑛𝑡 𝑎𝑟𝑚 𝑜𝑓 𝑡ℎ𝑖𝑠 𝑓𝑜𝑟𝑐𝑒𝑠 𝑖𝑠 𝑧𝑒𝑟𝑜),

$M_{P_1}=-P_1\cdot \frac L2 cos\alpha$ (𝑚𝑖𝑛𝑢𝑠 𝑠𝑖𝑔𝑛 𝑏𝑒𝑐𝑎𝑢𝑠𝑒 𝑜𝑓 𝑡ℎ𝑒 𝑑𝑖𝑟𝑒𝑐𝑡𝑖𝑜𝑛 𝑜𝑓 𝑓𝑜𝑟𝑐𝑒)

$M_{P_2}=-P_2\cdot dcos\alpha,$

$M_{F_{fr2}}=F_{fr2}\cdot Lcos\alpha,$

$M_{N_2}=N_2\cdot Lsin\alpha,$

→ (6):

$F_{fr2}\cdot Lcos\alpha+N_2\cdot Lsin\alpha-P_2\cdot dcos\alpha-P_1\cdot \frac L2 cos \alpha=0,$

$d=\frac{1}{P_2 cos\alpha}\cdot(F_{fr2}\cdot Lcos\alpha+N_2\cdot Lsin\alpha-P_1\cdot \frac L2 cos \alpha)=8.84~\text{m}.$