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A uniform beam, [AB], lies horizontally and in equilibrium on supports at points C and D of the beam, as shown in the diagram - Leaving Cert Applied Maths - Question 7 - 2019
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A uniform beam, [AB], lies horizontally and in equilibrium on supports at points C and D of the beam, as shown in the diagram. The mass of the beam is 12 kg and the... show full transcript
Worked Solution & Example Answer:A uniform beam, [AB], lies horizontally and in equilibrium on supports at points C and D of the beam, as shown in the diagram - Leaving Cert Applied Maths - Question 7 - 2019
Step 1
(i) Find the reaction forces at C and D
Answer
To find the reaction forces at points C and D, we first calculate the weight of the beam:
- Weight of beam, W = mass × gravity = 12 kg × 9.81 m/s² = 117.72 N.
Using the principles of equilibrium, we can set up the moments about point C (taking clockwise moments as positive). The moment due to the weight of the beam is:
- Moment due to weight of beam = W × distance from C to the center of the beam = 117.72 N × 35 cm = 41.2 Nm.
Let the reaction forces at C and D be R₁ and R₂, respectively. Summing moments about point C gives us:
- R₂ × 20 cm = 41.2 Nm
From this, we can calculate R₂:
Next, using the balance of vertical forces:
- R₁ + R₂ = 117.72 N
Substituting for R₂:
R₁ = 117.72 - 2.06 = 115.66 N$$ Thus, the reaction forces are: - R₁ = 115.66 N - R₂ = 2.06 NStep 2
(ii) A mass of m kg is now placed at B. This causes the beam to be on the point of lifting off the support at C. Find the value of m.
Answer
When the mass m is placed at B, the beam will tend to rotate about point C, making R₁ = 0. Using the moment balance about point C again:
The moment due to the mass at B:
- m × g × distance from C to B = m × 9.81 × 70 cm = m × 9.81 × 0.7 m.
Setting this equal to the moment due to R₂ about point C:
- R₂ × 20 cm = m × 9.81 × 0.7
Substituting the value of R₂ from question (i):
- 2.06 N × 20 cm = m × 9.81 × 0.7
Calculating:
- 2.06 × 0.2 = m × 9.81 × 0.7
- 0.412 = m × 6.867
- m = ( \frac{0.412}{6.867} ) = 0.060 kg
Therefore, the value of m is approximately 0.060 kg.
Step 3
Find the coefficient of friction between the ladder and the ground.
Answer
To find the coefficient of friction (μ) between the ladder and the ground, analyze the equilibrium and forces acting on the ladder.
The vertical forces must balance:
- R = 180 N (Weight of the ladder)
- ( R = \mu \cdot R_ \)
Considering the moments about the point of contact with the ground:
-
Using trigonometry:
- Height = 6 m × sin(θ)
- Distance to wall = 6 m × cos(θ)
-
The moments equation around the base:
- ( R_1 × 6 m × \sin(θ) = 180 × [6\cos(θ)] )
From the equilibrium conditions and substituting values:
- At point of slipping, we rearrange to find μ.
- Substituting known values: [ R_1 = 26.25 N \text{ and } R = 180 N ]
dives us: [ μ = \frac{26.25}{180} = 0.1458 ]
Thus, the coefficient of friction (μ) is approximately 0.1458.