Engineering Mechanics - Friction - Discussion

Discussion :: 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.

[A]. x = 0.678 m
[B]. x = 0.508 m
[C]. x = 1.005 m
[D]. x = 0.712 m

Answer: Option C

Explanation:

No answer description available for this question.

Sheshadri said: (Feb 21, 2011)  
I am not underderstand please explain.

Shubham said: (Mar 5, 2011)  
Please clearify the solution in a easy manner.

Thanks.

Raveesha said: (Jun 7, 2011)  
Please give the solution.

Laxman said: (Feb 26, 2015)  
Please solve this problem step by step sir.

Ani... said: (May 28, 2015)  
Pleas give the complete explanation briefly.

Filip said: (Jul 22, 2015)  
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.

Shyam said: (Sep 9, 2015)  
I can't understand this problem please explain simple logically.

Nikhil said: (Dec 30, 2016)  
I did not understand. Please explain it.

Raj said: (Sep 18, 2019)  
Well said, thanks @Filip.

Raj said: (Sep 18, 2019)  
Well said, thanks @Filip.

Raj said: (Sep 18, 2019)  
Well said, thanks @Filip.

Pranavjathrey said: (Aug 10, 2020)  
@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.

Jayanta said: (Mar 13, 2021)  
@Pranavjathrey.

Well done, thanks.

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