Question

A 5.00 m long ladder leans against a frictionless wall. The point of constant between the ladder and the wall is 4.00 m above the ground. The ladder is uniform with a mass of 15.00 kg. Determine the forces exerted by the ground and the wall on the ladder.

Answer 1

Answer on Question #62706, Physics / Other

A $5.00\mathrm{m}$ long ladder leans against a frictionless wall. The point of constant between the ladder and the wall is $4.00\mathrm{m}$ above the ground. The ladder is uniform with a mass of $15.00\mathrm{kg}$ . Determine the forces exerted by the ground and the wall on the ladder.

Solution:

$$\sin \theta = \frac {4 m}{5 m}$$ $$\cos \theta = \frac {3 m}{5 m}$$

Put pivot at bottom left end to get the normal force by the wall $N$ from torque:

$$(N) (4 m) = (m g) (2. 5 m) (\cos \theta)$$ $$N = \frac {(m g) (2 . 5 m) (\cos \theta)}{4 m} = \frac {(1 5 . 0 0 k g) (9 . 8 0 m / s ^ {2}) (2 . 5 m) (\frac {3}{5})}{4 m} = 5 5. 1 2 5 N$$

Use net force to get $F_{x}$ and $F_{y}$ :

$$F _ {n e t x} = 0 \quad F _ {x} - N = 0$$ $$F _ {x} = 5 5. 1 2 5 \mathrm {N}$$ $$F _ {n e t y} = 0 \quad F _ {y} - m g = 0$$ $$F _ {y} = m g = (1 5. 0 0 k g) \left(9. 8 0 \frac {m}{s ^ {2}}\right) = 1 4 7 \mathrm {N}$$

Answer: $N = 55.125 \, \text{N}$ ; $F_{y} = 147 \, \text{N}$ .

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