A plank AB has mass 40 kg and length 3 m. A load of mass 20 kg is attached to the plank at B. The loaded plank is held in equilibrium, with AB horizontal, by two vertical ropes attached at A and C, as shown in Figure 1. The plank is modelled as a uniform rod and the load as a particle. Given that the tension in the rope at C is three times the tension in the rope at A, calculate (a) the tension in the rope at C; (b) the distance CB.

A-Level Edexcel Maths Mechanics Moments: A plank AB has mass 40 kg and le

A plank AB has mass 40 kg and length 3 m - Edexcel - A-Level Maths Mechanics - Question 2 - 2005 - Paper 1

Question 2

A plank AB has mass 40 kg and length 3 m. A load of mass 20 kg is attached to the plank at B. The loaded plank is held in equilibrium, with AB horizontal, by two ver... show full transcript

Worked Solution & Example Answer:A plank AB has mass 40 kg and length 3 m - Edexcel - A-Level Maths Mechanics - Question 2 - 2005 - Paper 1

Step 1

(a) the tension in the rope at C;

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Answer

To find the tension in the rope at C, we begin by applying the principle of equilibrium. The equation of the forces in the vertical direction can be expressed as:

$T + 3T = 40g + 20g$

where:

$T$ is the tension in the rope at A,
$3T$ is the tension in the rope at C,
$40g$ is the weight of the plank (40 kg),
$20g$ is the weight of the load (20 kg).

Combining the terms:

$4T = 60g$

Solving for $T$ :

$T = \frac{60g}{4} = 15g$

Substituting the value of $g$ (approximately 9.81 m/s²):

$T = 15 \times 9.81 \approx 147.15 N$

Therefore, the tension at C is:

$3T = 3 \times 15g = 45g$

Calculating this:

$3T \approx 441 N$

Thus, the tension in the rope at C is approximately 441 N.

Step 2

(b) the distance CB.

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Answer

To find the distance CB, we need to consider the moments about point B. The equation for moments can be given as:

$15g \times 3 = 40g \times d$

where:

$15g$ is the force acting at C,
$3$ m is the distance from A to B,
$40g$ is the weight of the plank,
$d$ is the distance from B to C.

Rearranging the equation gives:

$d = \frac{15g \times 3}{40g} = \frac{45}{40} = 1.125 m$

Subtracting this from the total length 3 m:

$CB = 3 - d = 3 - 1.125 = 1.875 m$

Therefore, the distance CB is approximately 1.875 m.

A plank AB has mass 40 kg and length 3 m - Edexcel - A-Level Maths Mechanics - Question 2 - 2005 - Paper 1

Worked Solution & Example Answer:A plank AB has mass 40 kg and length 3 m - Edexcel - A-Level Maths Mechanics - Question 2 - 2005 - Paper 1

(a) the tension in the rope at C;

(b) the distance CB.

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