Angle of Inclination Homework: Find θ at Ladder's Verge of Slipping

[SpiraRoam]

Homework Statement

A ladder of length 5 m is resting against a SMOOTH wall, its point of contact being just below a first floor window. The bottom of the ladder rests on the ground and the coefficient of friction between the ladder and the ground is 0.28

A person of mass 78 kg climbs the ladder, which has a mass of 6 kg. The bottom of the ladder is at the point of slipping, outward from the wall, just as the person reaches the mid-point of the ladder’s length.

Determine the angle of inclination (θ) of the ladder, when the person is at the mid-point and the ladder is on the verge of slipping.

Homework Equations

I'm not sure whether the angle of repose is relevant? θ=tan^-1μ with μ being the coefficient of friction

w=mg and N (normal reaction) =mg

The Attempt at a Solution

I was going to apply the method from exercise 13.1.3 at http://physique.merici.ca/mechanics/chap13mech.pdf

But is this relevant to to the angle at the verge of slipping? I can't see how the angle of repose is the answer - the question is 6 marks and that's too simple of a solution

Thanks

 

[kuruman]

Please show your attempt at the solution including a free body diagram that applies to this problem. The angle of repose may or may or may not be relevant. Let's see what comes out of the calculation.

 

[SpiraRoam]

Forces: X Y Torque
Weight of ladder 0 -58.86N x 5m x sin(90°- Ɵ)
Weight of Person 0 -765.18N 765.18N x 5m x sin(90°-ϴ)
Normal force wall FN1 0 0 Nm
Normal force ground 0 FN2 0 Nm
Friction force ground -Ff 0 0 Nm

That was supposed to be a table ^, but I can't seem to paste it here.

I would then put the results into the mechanics equations:

ΣF_x = F_N1 - F_f = 0
ΣF_y = -98N-245N + F_N2 = 0
Στ = -294.93Nm x sin(90 - θ) - 3825.9Nm x sin(90 - θ) + F_N1 x 5m x sin(θ)=0
= -294.93 Nm x cos(90) - 3825.9 Nm cos(90) + F_N1 x 5m x sinθ
= 4120.83 Nm x cosθ = F_N1 x 5m x sinθ
= 4120.83 / 5 = F_N1 x tanθ
= 824.166 = F_f x tanθ

824.166N ≤ μF_N2 x tanθ
824.166N ≤ 0.28 x F_N2 x tanθ
824.166N ≤ 0.28 x 824.04 x tanθ
824.166N / 230.7312 ≤ tanθ
3.572 ≤ tanθ
tan^-1(3.572) ≤ θ
θ = 74.36°

Thanks

 

[kuruman]

θ = 74.36°

That's not the number I got. It's difficult for me to check your work with all those numbers thrown in, but I suspect your torque equation is incorrect. If you repost, please use a symbolic expression in the form tanθ = a function of the relevant quantities, i.e. m_person, m_ladder, length of ladder, μ and g.

 

[SpiraRoam]

Not sure where I'd find / derive such an expression, there's no hint of it in the course notes..we've only been given the ones I mentioned

 

[haruspex]

Weight of ladder 0 -58.86N x 5m x sin(90°- Ɵ)

The ladder is 5m long. How far along it is the mass centre? Likewise, reconsider the torque for the man.

As kuruman suggests, it is far better to keep everything symbolic. Create symbols for all the given data, like W_L for the weight of the ladder, etc. Do not plug in numbers until the end. This has many advantages, both for you and for the reader. They should teach you this.

 

[SpiraRoam]

Right in the middle, so 2.5m along

Would the angle of repose be subtracted from the final angle? Or is it not relevant? I'm getting a final value of 82.032 degrees

 

[haruspex]

Right in the middle, so 2.5m along

So why is that not in your calculation of torque?

 

[SpiraRoam]

It is, I've just corrected it to get the final value of 82.032 degrees - thanks for the heads up =]

 

[haruspex]

It is, I've just corrected it to get the final value of 82.032 degrees - thanks for the heads up =]

Ok.
But to make your work intelligible, please repost using W for the weight total (the two weights are at the same point, so no need to distinguish), L for length of ladder, F for the frictional force and μ for the coefficient. If handling two normal forces, distinguish with a subscript.
When you write a torque equation, you should state your chosen axis.

It asks for the inclination of the ladder. That suggests to me they want the angle to the vertical, but I could be wrong.