A uniform beam has a mass of 18 kg and rests on two surfaces at points A and B. Determine the maximum distance x to which the girl can slowly walk up the beam before it begins to slip. The girl has a mass of 50 kg and walks up the beam with a constant velocity.

 A uniform beam has a mass of 18 kg and rests on two surfaces at points A and B. Determine the maximum distance x to which the girl can slowly walk up the beam before it begins to slip. The girl has a mass of 50 kg and walks up the beam with a constant velocity.

The solution is;

Let θ be the internal angle at B
sinθ = 1.8/3.0
sinθ = 0.6
θ = 36.87.

At B, the normal force of the floor on the ramp Nb is vertical.
At A, the normal force of the step on the ramp Na is perpendicular to the ramp or θ from vertical.

Sum horizontal forces to zero. Assume right is the positive direction.

Na(sinθ) - μ(Na)cosθ - μNb = 0.
Na(sinθ - μcosθ) = μNb.
Nb = Na(0.6 - 0.2(0.8)/0.2.
Nb = 2.2Na.

Sum vertical forces to zero. Assume up is positive.

Nb + Na(cosθ) + μ(Na)sinθ - (M + m)g = 0.
2.2Na + Na(cosθ) + μ(Na)sinθ - (M + m)g = 0,
Na (2.2 + cosθ + μsinθ) = (M + m)g,
Na = (M + m)g / (2.2 + cosθ + μsinθ),
Na = (50 + 18)9.81 / (2.2 + 0.8 + 0.2(0.6)),
Na = 213.8.

Sum moments about B to zero. Assume clockwise moment is positive'.

Nb and the friction force at A have no moment arm so create no moment about B.
Na[3.0] - Mg[xcosθ] - mg[(3.5 / 2)cosθ] = 0.

x = (3Na - mg[1.75cosθ])/Mgcosθ.
x = (3(213.8) - 18(9.81)1.75(0.8))/50(9.81)0.8.
x = 1.005 m.