Question

How can we solve this one?

Answer 1

First, we have to establish a relationship between the angle of inclination of the incline $\sf (\theta)$ and the length of the ramp $\sf (L).$

Refer to the attachment.

$\sf sin(\theta) = \dfrac{1}{L}$

$\longrightarrow \sf L = \dfrac{1}{sin(\theta)}\quad\quad \dots(1)$

Refer to the attachment for F.B.D.

According to the question, "..... is needed to slide the crate up the ramp at a constant speed...."

Constant speed means, $\sf a = 0\:m/s^2\quad \quad \dots(2)$

According to the F.B.D.,

$\sf (400) - \{mg\:sin(\theta)\} = ma$

From (2)

$\longrightarrow \sf (400) - \{mg\:sin(\theta)\} = m\times 0$

$\longrightarrow \sf 400 = mg\:sin(\theta)$

Putting $\sf m = 60\:kg,$

$\longrightarrow \sf 400 = 60 \times 9.8 \times sin(\theta)$

$\longrightarrow \sf sin(\theta) = \dfrac{400}{60\times9.8}$

$\longrightarrow \sf sin(\theta) = 0.680\quad\quad \dots (3)$

Putting the value of (3) in (1),

$\sf L = \dfrac{1}{sin(\theta)}$

$\longrightarrow \sf L = \dfrac{1}{0.680}$

$\longrightarrow \underline{\underline{\green{\sf{ L = 1.47\:m }}}}$