A plank AB has mass 12 kg and length 2.4 m. A load of mass 8 kg is attached to the plank at the point C, where AC = 0.8 m. The loaded plank is held in equilibrium, with AB horizontal, by two vertical ropes, one attached at A and the other attached at B, as shown in Figure 2. The plank is modelled as a uniform rod, the load as a particle and the ropes as light inextensible strings. (a) Find the tension in the rope attached at B. (4) The plank is now modelled as a non-uniform rod. With the new model, the tension in the rope attached at A is 10 N greater than the tension in the rope attached at B. (b) Find the distance of the centre of mass of the plank from A. (6)

A-Level Edexcel Maths Mechanics Moments: A plank AB has mass 12 kg and le

A plank AB has mass 12 kg and length 2.4 m - Edexcel - A-Level Maths Mechanics - Question 6 - 2008 - Paper 1

Question 6

A plank AB has mass 12 kg and length 2.4 m. A load of mass 8 kg is attached to the plank at the point C, where AC = 0.8 m. The loaded plank is held in equilibrium, w... show full transcript

Worked Solution & Example Answer:A plank AB has mass 12 kg and length 2.4 m - Edexcel - A-Level Maths Mechanics - Question 6 - 2008 - Paper 1

Step 1

Find the tension in the rope attached at B.

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Answer

To find the tension in the rope attached at B, we first establish a moment about point A to sum the moments equals zero.

Calculate the total moment due to the weights acting on the plank:
- The weight of the plank acts at its center,
  
  \text{Moment due to plank} = 12g imes 1.2, ext{m}
where $g$ = 9.81 m/s² (approx. 10 m/s² can also be used for simplicity).
The moment due to the load is:
- Moment due to the load at C:
\text{Moment due to load} = 8g imes 0.8, ext{m}
Setting the moments around point A to zero gives us:

$12g imes 1.2 + 8g imes 0.8 = X imes 2.4$

where $X$ is the tension at point B.
By solving the equation:

$X = \frac{(12g \times 1.2 + 8g \times 0.8)}{2.4}$

Insert values to calculate:

$X = \frac{(12 \times 9.81 \times 1.2 + 8 \times 9.81 \times 0.8)}{2.4} \approx 85\, ext{N}$

Step 2

Find the distance of the centre of mass of the plank from A.

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Answer

To find the distance of the center of mass of the plank from A when it is modeled as a non-uniform rod:

From the problem statement, we know the tension at A is $X + 10$ .
Set up the moment equation again around point R (which balances the total forces):

$(X + 10) \times x + X \times 0.8 = 8g \times 0.8 + 12g \times 1.2$
We can solve this equation for $x$ , which represents the distance from A to the center of mass:

\text{Total moments should balance out:} \$8g \times 0.8 + 12g \times 1.2 = R, ext{which in this case is } 93, ext{N.}

By simplifying this equation:

$x = \frac{(X + 10) \times x + X \times 0.8}{(8g + 12g)}$
Substitute known values to solve for $x$ :

After solving, we find that: $x \approx 1.4\, ext{m}$

Hence, the center of mass of the plank is approximately 1.4 m from point A.

A plank AB has mass 12 kg and length 2.4 m - Edexcel - A-Level Maths Mechanics - Question 6 - 2008 - Paper 1

Worked Solution & Example Answer:A plank AB has mass 12 kg and length 2.4 m - Edexcel - A-Level Maths Mechanics - Question 6 - 2008 - Paper 1

Find the tension in the rope attached at B.

Find the distance of the centre of mass of the plank from A.

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