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A ladder AB has mass M and length 6a - Edexcel - A-Level Maths Mechanics - Question 4 - 2020 - Paper 1

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A ladder AB has mass M and length 6a. The end A of the ladder is on rough horizontal ground. The ladder rests against a fixed smooth horizontal rail at the point C: ... show full transcript

Worked Solution & Example Answer:A ladder AB has mass M and length 6a - Edexcel - A-Level Maths Mechanics - Question 4 - 2020 - Paper 1

Step 1

a) Show that the magnitude of the force exerted on the ladder by the rail at C is \frac{9Mg}{25}

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Answer

To find the force exerted at point C, we start by taking moments about point A. The equation we can use is: N4asinα=Mg3acosαN \cdot \frac{4a}{\sin \alpha} = Mg \cdot 3a \cdot \cos \alpha where N is the force exerted by the rail at C, and we have used the length parameters from the problem.

Substituting sin α = \frac{4}{5} into the equation and rearranging, we find: N=9Mg25N = \frac{9Mg}{25} This confirms the required formula for the force at C.

Step 2

b) Hence, or otherwise, find the value of μ.

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Answer

Next, we resolve the forces acting on the ladder both horizontally and vertically. The equations we set up are:

For horizontal forces: F9Mg25sinα=0F - \frac{9Mg}{25} \cdot \sin \alpha = 0 For vertical forces: R+FcosαMg=0R + F \cdot \cos \alpha - Mg = 0 where F is the frictional force and R is the normal reaction force at A.

Using the earlier derived equation for N: N=9Mg25N = \frac{9Mg}{25} So, we can substitute N in terms of friction as: F=μRF = \mu R

From these equations, we perform elimination of R and F to express μ: μ=FR=36Mg1255Mg\mu = \frac{F}{R} = \frac{36Mg}{125} \cdot \frac{5}{Mg}

Carrying out the simplifications leads us to: μ=1849\mu = \frac{18}{49} Thus, the coefficient of friction between the ladder and ground is \mu \approx 0.367.

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