Moments Problem

[jonnburton]

Hi,

I was wondering if anyone could help me with this problem which I am not sure how to approach.

The diagram (copied below), shows a man of mass 70kg at rest while abseiling down a vertical cliff. Assume that the rope is attached to the man at his centre of mass. You should model the man as a uniform rod and assume that he is not holding the rope.

a) Draw a diagram to show the forces acting on the man



b)Find the minimum value of the coefficient of friction between his feet and the cliff if he remains at rest.


I do find these problems quite hard to visualise.

As far as I can see, the reaction force S and the weight (686N) are trying to rotate around point A clockwise. The tension in the rope is trying to rotate around point A anticlockwise.


If, for the sake of this we can take the distance from the rope to the feet as 1 metre:

The perpendicular distance of the weight force from A is cos 30, so the moment of this force around A is 686cos30

As the system is in equilibrium, T opposes this rotation, so its moment is also 686cos30 (in the opposite direction).

But I need to get the value for the frictional force and R... I think the problem is that I'm failing to visualise what needs to be done here and I'd be very grateful if anyone could tell me how to approach the problem.

 

[DrPhil]

Hi,

I was wondering if anyone could help me with this problem which I am not sure how to approach.

The diagram (copied below), shows a man of mass 70kg at rest while abseiling down a vertical cliff. Assume that the rope is attached to the man at his centre of mass. You should model the man as a uniform rod and assume that he is not holding the rope.

a) Draw a diagram to show the forces acting on the man

View attachment 3080

b)Find the minimum value of the coefficient of friction between his feet and the cliff if he remains at rest.


I do find these problems quite hard to visualise.

As far as I can see, the reaction force S and the weight (686N) are trying to rotate around point A clockwise. The tension in the rope is trying to rotate around point A anticlockwise.


If, for the sake of this we can take the distance from the rope to the feet as 1 metre:
Lets make that a parameter, L, instead of assuming a value. We need to keep track of UNITS.

The perpendicular distance of the weight force from A is cos 30, so the moment of this force around A is 686cos30
...............................................................L cos(30°)................................(686 N)L cos(30°)

As the system is in equilibrium, T opposes this rotation, so its moment is also 686cos30 (in the opposite direction).

But I need to get the value for the frictional force and R... I think the problem is that I'm failing to visualise what needs to be done here and I'd be very grateful if anyone could tell me how to approach the problem.

I usually start by decomposing the Weight force into components parallel and perpendicular to the angles in the problem. These components are balanced by rope tension \(\displaystyle T\) and the force of the cliff against his feet. You need a name for that Force .. we can use \(\displaystyle D\), which acts in the direction from A to B (the rigid rod pushes and the rope tension pulls).

\(\displaystyle T = W\ \cos\theta = (686\ \mathrm N)\cos (30°)\)
\(\displaystyle D = W\ \sin\theta = (686\ \mathrm N)\sin (30°)\)

The three forces acting on the man are \(\displaystyle W\), \(\displaystyle T\), and \(\displaystyle D\).

Now consider the forces at point A. [Don't think we have to use moments!?] The force \(\displaystyle D\) has a downward component that must be balanced by friction, \(\displaystyle F\), and a normal component \(\displaystyle N\). At the point where slippage might start,

\(\displaystyle F = \mu\ N \implies \mu = F/N = \cdot \cdot \cdot \)

Easier than you thought!!

Is your force \(\displaystyle R\) supposed to be what I called \(\displaystyle D\)?
At point C, the net force can't have a component perpendicular to the rope - so \(\displaystyle S\) may be spurious.

 

[jonnburton]

I usually start by decomposing the Weight force into components parallel and perpendicular to the angles in the problem. These components are balanced by rope tension \(\displaystyle T\) and the force of the cliff against his feet. You need a name for that Force .. we can use \(\displaystyle D\), which acts in the direction from A to B (the rigid rod pushes and the rope tension pulls).

\(\displaystyle T = W\ \cos\theta = (686\ \mathrm N)\cos (30°)\)
\(\displaystyle D = W\ \sin\theta = (686\ \mathrm N)\sin (30°)\)

The three forces acting on the man are \(\displaystyle W\), \(\displaystyle T\), and \(\displaystyle D\).

Now consider the forces at point A. [Don't think we have to use moments!?] The force \(\displaystyle D\) has a downward component that must be balanced by friction, \(\displaystyle F\), and a normal component \(\displaystyle N\). At the point where slippage might start,

\(\displaystyle F = \mu\ N \implies \mu = F/N = \cdot \cdot \cdot \)

Easier than you thought!!

Is your force \(\displaystyle R\) supposed to be what I called \(\displaystyle D\)?
At point C, the net force can't have a component perpendicular to the rope - so \(\displaystyle S\) may be spurious.

Thank you very much for explaining that, DrPhil. I have been going through it. At one point I couldn't see why \(\displaystyle T = W cos\theta\) and \(\displaystyle D = W sin \theta\), but after drawing a separate forces triangle, this became clear.

I still need to practice moments quite a lot because they aren't obvious to me. One other thing I learnt here was that forces diagrams are totally separate from the diagram depicting the problem in terms of measurements...

The R and the S on my diagram were my mistake. I was under the impression that these would be normal reaction forces.