A ladder 20 − 𝑚. long is leaning against an embankment inclined 60o with the horizontal.

Question

A ladder 20 − 𝑚. long is leaning against an embankment inclined 60^o with the horizontal. If the bottom of the ladder is being moved horizontally towards the embankment at 1 𝑚⁄𝑠 , how fast is the top of the ladder moving when the bottom is 4 𝑚. from the embankment?

Hint: Cosine Law: 𝑐² = 𝑎² + 𝑏² − 2𝑎𝑏 cos C

Answer 1

I am not able to provide a picture, so please try to draw the situation.

Let

A be the point where the ladder touches the ground,

B be the point where the ladder touches the wall,

C be the point where the wall touches the ground,

x be the distance AC (along the ground),

y be the distance BC (along the wall).

We are given

L = 20 m is the length of the ladder, i.e., the distance AB,

the angle ACB = 180° - 60° = 120°,

dx/dt = -1 m/s

(Note that dx/dt is negative because x is shrinking.)

We are to calculate dy/dt.

By the cosine theorem

L² = x² + y² - 2xycos(120°)

Using cos(120°) = -1/2

L² = x² + y² + xy (1)

Differentiating (1) with respect to time

0 = 2xdx/dt + 2ydy/dt + xdy/dt + ydx/dt

Substituting the give dx/dt = -1 into the above

0 = -2x + 2ydy/dt + xdy/dt - y

Expressing dy/dt from the above

dy/dt = (2x + y)/(x + 2y) (2)

We are to calculate dy/dt when x = 4m.

For that we first need to calculate y when x = 4m.

We do that from (1)

20² = 4² + y² + 4y

This is a quadratic equation to be solved

y² + 4y - 384 = 0

y = (-4 ±√(4² + 4×384))/2 = ±√388 - 2

Of the two solutions only √388 - 2 is physically feasible.

Now we can substitute the known x and y into (2)

dy/dt = (2×4 + √388 - 2)/(4 + 2(√388 - 2)) = 0.62 m/s