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I was given the following task:

enter image description here

My problem is the following:

For $A$ and $B$ I’ve got friction forces $R_A$ and $R_B$. I can calculate those by using the formula: $R_A = N_A \cdot \mu $
$\mu $ is given with $0.5$

Now let‘s say the normal force on point $A$ is really high, according to the formula, the friction force should be really high aswell. This does not seem logical for me. Shouldn‘t the friction be only there if the ladder is being push on the ground?

How are those tasks usually calculated?

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    $\begingroup$ $N_A\cdot \lambda$ is the maximum friction force. If the ladder is sliding, the friction force will have this value. When the ladder is standing the still, the force will be lower than this value. $\endgroup$ – Crimson Jan 16 '18 at 14:02
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    $\begingroup$ The problem is solved by stating that for equilibrium both the sum of the forces and the sum of the torques (wrt an arbritrary axis) are zero. $\endgroup$ – Crimson Jan 16 '18 at 14:04
  • $\begingroup$ I will answer this question tomorrow, time permitting. $\endgroup$ – Gert Jan 16 '18 at 20:34
  • $\begingroup$ Actually I just found the solution a few minutes ago. But thank you!:) $\endgroup$ – Finn Eggers Jan 16 '18 at 20:35
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Shouldn‘t the friction be only there if the ladder is being push on the ground?

Static friction is present if

  • There is no relative motion

and

  • The point of contact would experience relative motion in the absence of friction

The problem asks you to assume the former (we're calculation the limits of the non-slipping situation, right?), and it is certainly the case that the ladder would slide at both ends if there were no frictional forces at all.

So, you need to account for the static fiction at both ends.

How are those tasks usually calculated?

The key observation here is that in the cases we care about the ladder is still—not moving, much less accelerating. So you can approach this as a static equilibrium problem (including a force from the trainee in part 2).

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