Question

A 150-kg ladder leans against a smooth wall, making an angle of 30 degrees with the floor. The centre of gravity of the ladder is one-third the way up from the bottom. How large a horizontal force must the floor provide if the ladder is not to slip?

Answer 1

Answer on Question #41193 – Physics – Mechanics

A 150-kg ladder leans against a smooth wall, making an angle of 30 degrees with the floor. The centre of gravity of the ladder is one-third the way up from the bottom. How large a horizontal force must the floor provide if the ladder is not to slip?

Solution:

$\alpha = 30{}^{\circ}$ – angle which ladder makes with the wall

$\mathrm{N}_{\mathrm{A}}$ – reaction force from the floor

$\mathrm{N}_{\mathrm{B}}$ – reaction force from the wall

$\mathrm{m} = 150 \mathrm{~kg} - \mathrm{mass}$ of the ladder

L – length of the ladder

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

$$\vec {\mathrm {F}} + \mathrm {m} \vec {\mathrm {g}} + \overrightarrow {\mathrm {N} _ {\mathrm {A}}} + \overrightarrow {\mathrm {N} _ {\mathrm {B}}} = \vec {0}$$

Projection of the law on the X-axis:

$$x: N _ {B} - F = 0 \Rightarrow N _ {B} = F$$

Momentum equation for point A (the second law of equilibrium):

$$\mathrm {A}: \mathrm {M} _ {\mathrm {m g}} + \mathrm {M} _ {\mathrm {F}} + \mathrm {M} _ {\mathrm {N} _ {\mathrm {A}}} + \mathrm {M} _ {\mathrm {N} _ {\mathrm {B}}} = 0$$

$(\mathrm{M}_{\mathrm{N}_{\mathrm{A}}} = \mathrm{M}_{\mathrm{F}} = 0,$ because moment arm of this forces is zero)

$\mathrm{M}_{\mathrm{mg}} = -\mathrm{mg} \cdot \frac{\mathrm{L}}{3} \cos \alpha$ (minus sign because of the direction of force)

$$\begin{array}{l} \mathrm{M}_{\mathrm{N}_{\mathrm{B}}} = \mathrm{N}_{\mathrm{B}} \cdot \mathrm{L} \sin \alpha = |\mathrm{using} (1)| = \mathrm{F} \cdot \mathrm{L} \sin \alpha \\ \rightarrow (3): \end{array}$$ $$\mathrm{F} \cdot \mathrm{L} \sin \alpha - \mathrm{mg} \cdot \frac{\mathrm{L}}{3} \cos \alpha = 0$$ $$\mathrm{F} = \frac{\mathrm{mg}}{3} \cdot \frac{\cos \alpha}{\sin \alpha} = \frac{\mathrm{mg}}{3 \tan \alpha} = \frac{\mathrm{mg}}{3 \tan \alpha} = \frac{150 \mathrm{~kg} \cdot 9.8 \frac{\mathrm{N}}{\mathrm{kg}}}{3 \cdot \tan 30{}^{\circ}} = 849 \mathrm{~N}$$

Answer: horizontal force that the floor provide is equal to 849N.

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