A uniform beam AB has mass 20 kg and length 6 m. The beam rests in equilibrium in a horizontal position on two smooth supports. One support is at C, where AC = 1 m, and the other is at the end B, as shown in Figure 1. The beam is modelled as a rod. (a) Find the magnitudes of the reactions on the beam at B and C. A boy of mass 30 kg stands on the beam at the point D. The beam remains in equilibrium. The magnitudes of the reactions on the beam at B and at C are now equal. The boy is modelled as a particle. (b) Find the distance AD.

A-Level Edexcel Maths Mechanics Forces: A uniform beam AB has mass 20 kg

A uniform beam AB has mass 20 kg and length 6 m - Edexcel - A-Level Maths Mechanics - Question 3 - 2011 - Paper 1

Question 3

A uniform beam AB has mass 20 kg and length 6 m. The beam rests in equilibrium in a horizontal position on two smooth supports. One support is at C, where AC = 1 m, ... show full transcript

Worked Solution & Example Answer:A uniform beam AB has mass 20 kg and length 6 m - Edexcel - A-Level Maths Mechanics - Question 3 - 2011 - Paper 1

Step 1

Find the magnitudes of the reactions on the beam at B and C.

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Answer

To determine the reactions at supports B and C, we need to take moments about point B to eliminate the unknown reaction at B.

Taking moments about B:

The total mass of the beam is 20 kg, which acts at its midpoint (3 m from A). The moment due to the weight of the beam about B is:

$ext{Moment from weight of beam} = 20 ext{ kg} imes 9.81 ext{ m/s}^2 imes 3 ext{ m} = 588.6 ext{ Nm}$

This is balanced by the reaction at C:

$R_C imes 5 ext{ m} = 588.6 ext{ Nm}$

Thus,

$R_C = \frac{588.6}{5} = 117.72 ext{ N}$
1. Resolving vertically:
Since the system is in equilibrium, all vertical forces must balance:

$R_C + R_B = 20 imes 9.81$

Substituting for $R_C$ :

$117.72 + R_B = 196.2 ext{ N}$

Solving for $R_B$ gives:

$R_B = 196.2 - 117.72 = 78.48 ext{ N}$

Step 2

Find the distance AD.

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Answer

To find the distance AD when the boy stands at point D:

Resolving the forces vertically:

The weight of the boy adds an extra force downwards, hence:

$50 ext{ N} = R + R_C$

Where $R$ is the reaction at B and is also equal to the reaction at C since they are now equal.
1. Taking moments about B again:
The boy's weight acts at D, with distance AD = x. Setting up the moment equation gives:

$30 ext{ kg} imes 9.81 ext{ m/s}^2 imes (6 - x) = 3 imes 20 ext{ kg} imes 9.81$

Simplifying:

$294 imes (6 - x) = 588$

Solving:

$1764 - 294x = 588$

$294x = 1176$

$x = \frac{1176}{294} = 4 ext{ m}$

Thus, the distance AD is then calculated as:

$AD = 1 ext{ m} + x = 1 + 4 = 5 ext{ m}$

A uniform beam AB has mass 20 kg and length 6 m - Edexcel - A-Level Maths Mechanics - Question 3 - 2011 - Paper 1

Worked Solution & Example Answer:A uniform beam AB has mass 20 kg and length 6 m - Edexcel - A-Level Maths Mechanics - Question 3 - 2011 - Paper 1

Find the magnitudes of the reactions on the beam at B and C.

Find the distance AD.

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