Engineering Mechanics - Friction - Discussion
Discussion Forum : Friction - General Questions (Q.No. 2)
2.
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.
Discussion:
13 comments Page 1 of 2.
Jayanta said:
4 years ago
@Pranavjathrey.
Well done, thanks.
Well done, thanks.
Pranavjathrey said:
5 years ago
@All
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.
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.
Raj said:
6 years ago
Well said, thanks @Filip.
Raj said:
6 years ago
Well said, thanks @Filip.
Raj said:
6 years ago
Well said, thanks @Filip.
Nikhil said:
9 years ago
I did not understand. Please explain it.
Shyam said:
10 years ago
I can't understand this problem please explain simple logically.
Filip said:
1 decade ago
1st: draw the fbd.
2nd: Summation of Fx and Fy.
3rd: We now have 2 unknowns; Na and Nb.
4th: Use your calculator to get the unknowns.
5th: Sum of moment at B using the beam as the x-axis.
Answer: 0.678 m.
2nd: Summation of Fx and Fy.
3rd: We now have 2 unknowns; Na and Nb.
4th: Use your calculator to get the unknowns.
5th: Sum of moment at B using the beam as the x-axis.
Answer: 0.678 m.
Ani... said:
1 decade ago
Pleas give the complete explanation briefly.
Laxman said:
1 decade ago
Please solve this problem step by step sir.
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