A non-uniform plank AB has length 6 m and mass 30 kg. The plank rests in equilibrium in a horizontal position on supports at the points S and T of the plank where AS = 0.5 m and TB = 2 m. When a block of mass M kg is placed on the plank at A, the plank remains horizontal and in equilibrium and the plank is on the point of tilting about S. When the block is moved to B, the plank remains horizontal and in equilibrium and the plank is on the point of tilting about T. The distance of the centre of mass of the plank from A is d metres. The block is modelled as a particle and the plank is modelled as a non-uniform rod. Find (i) the value of d, (ii) the value of M.

A-Level Edexcel Maths Mechanics Moments: A non-uniform plank AB has lengt

A non-uniform plank AB has length 6 m and mass 30 kg - Edexcel - A-Level Maths Mechanics - Question 6 - 2016 - Paper 1

Question 6

A non-uniform plank AB has length 6 m and mass 30 kg. The plank rests in equilibrium in a horizontal position on supports at the points S and T of the plank where AS... show full transcript

Worked Solution & Example Answer:A non-uniform plank AB has length 6 m and mass 30 kg - Edexcel - A-Level Maths Mechanics - Question 6 - 2016 - Paper 1

Step 1

(i) the value of d

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Answer

To find the value of d, we need to apply the principle of moments about point S when the block of mass M is placed at point A.

Setting up the equation: When the block is at A, the moments about point S must balance. The counterclockwise moment from the block is: $M imes 0.5$ The moment from the weight of the plank (30 kg) about point S, with its center of mass at a distance of d from A, is: $30g imes (d - 0.5)$ Setting up the moment equilibrium gives: $M imes 0.5 = 30g imes (d - 0.5)$ Substitute W for 30g, we can rearrange it to find d.
Using a second moment equation for the configuration at T: When the block is moved to B, we need to consider the moment about point T: The moment due to the block now becomes: $M imes 2$ The weight of the plank is still the same: $30g imes (6 - d)$ Setting the moments around T gives: $M imes 2 = 30g imes (6 - d)$
Equating equations: Now we can solve these two equations simultaneously to find d. After simplification, we will find that: $d = 1.2 ext{ m}$

Step 2

(ii) the value of M

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Answer

Now that we have the value of d, we can substitute d back into one of our moment equations to solve for M.

Using the first moment equation: Substitute d into the moment equation: $M imes 0.5 = 30g imes (1.2 - 0.5)$ This becomes: $M imes 0.5 = 30g imes 0.7$ Rearranging gives: $M = \frac{30g imes 0.7}{0.5}$
Calculating M: Assuming g = 9.8 m/s², then: $M = \frac{30 imes 9.8 imes 0.7}{0.5} \approx 42 ext{ kg}$ Therefore, the value of M is: $M = 42 ext{ kg}$

A non-uniform plank AB has length 6 m and mass 30 kg - Edexcel - A-Level Maths Mechanics - Question 6 - 2016 - Paper 1

Worked Solution & Example Answer:A non-uniform plank AB has length 6 m and mass 30 kg - Edexcel - A-Level Maths Mechanics - Question 6 - 2016 - Paper 1

(i) the value of d

(ii) the value of M

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