Question

A 6.2 kg ladder, 1.97 m long, rests on two sawhorses. Sawhorse A is 0.64 m from one end of the ladder, and sawhorse B is 0.17 m from the other end of the ladder. What force does each sawhorse exert on the ladder?

Answer 1

Answer:

42.69 N and 18.07 N

Explanation:

We are given that

Mass of ladder=6.2 kg

Length of ladder=1.97 m

Distance of Sawhorse A from one end=0.64 m

Distance of sawhorse B from other end=0.17 m

Let center of Ladder=$\frac{1.97}{2}=0.985 m$

Now, the distance of sawhorse A from center=r=0.985-0.64=0.345 m

Distance of sawhorse B from center of ladder=0.985-0.17=0.815  m

Force one ladder due to gravity=mg=$6.2\times 9.8=60.76N$

Where $g=9.8 m/s^2$

Torque applied on Sawhorse A=$0.345F_a$

Torque applied on Sawhorse B=$0.815F_b$

In equilibrium

$0.345F_a=0.815F_b$

$F_b=\frac{0.345}{0.815}F_a$

Total force=$F_a+F_b$

$F_a+\frac{0.345}{0.815}F_a=60.76$

$\frac{0.815F_a+0.345F_a}{0.815}=60.76$

$\frac{1.16}{0.815}F_a=60.76$

$F_a=\frac{60.76\times 0.815}{1.16}=42.69 N$

$F_b=\frac{0.345}{0.815}\times 42.69=18.07 N$