In today’s entry, we will solve a mechanics problem involving a ladder laying on a peg. Specifically, we will play around with the ideas of moments in equilibrium and the limiting equilibrium.
Give the problem a try before jumping in for the solution!
Solution
By first drawing the diagram with forces annotated, we have the following
Since the object is in equilibrium, the sum of clockwise moments equals the sum of anti-clockwise moments.
We will now label the angle the ladder makes with the ground theta.
The component of weight which is perpendicular to the ladder is Wcos(theta).
Using trigonometry, we can find that
If we take moments about point A, we only need to consider the forces F1 and W.
Therefore,
We have found the answer to part A!
Now onto part B, we have to find the range of the coefficient of friction.
We are given in the question that the ladder is in equilibrium, we can make use of the following formula.
this means we have
We now have to find expressions of F2 and R using the fact that the object is in equilibrium.
We will first consider the vertical direction, it has a downward force of the weight, an upward force of the reaction as well as an upwards component of the force acting on the ladder by the peg.
Mathematically, we get
As for the horizontal direction, we have the frictional force F2 as well as a horizontal component of the force acting on the ladder by the peg.
We can now see that if we rearrange for F2 and R, we will have two expressions in terms of W and F1, where F1 is in terms of W.
and
Putting these two expressions into the inequality, we get
With a bit of algebra by multiplying the expression by 169 times 13, we will have
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Happy reading,
Barry 🍩