A uniform beam, AB, is held in a horizontal position by two vertical inelastic strings attached at the points C and D respectively. The weight of the beam is 20 N and the length of the beam is 4 m. A particle of weight 18 N is placed at point P on the beam and another particle of weight 10 N is placed at point Q on the beam. $|C| = |D| = 1 m$ and $|CP| = |QD| = \frac{1}{2} m$. Calculate the tension in each of the strings. A uniform ladder, of weight 160 N, is resting on rough horizontal ground and leaning against a smooth vertical wall. The length of the ladder is 8 m. The ladder makes an angle $\theta$ with the ground, where $\tan \theta = \frac{1}{2}$. The ladder is in equilibrium and is on the point of slipping. Find the coefficient of friction between the ladder and the ground.

Leaving Cert Applied Maths Statics: A uniform beam, AB, is held in a

A uniform beam, AB, is held in a horizontal position by two vertical inelastic strings attached at the points C and D respectively - Leaving Cert Applied Maths - Question 7 - 2015

Question 7

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A uniform beam, AB, is held in a horizontal position by two vertical inelastic strings attached at the points C and D respectively. The weight of the beam is 20 N a... show full transcript

Worked Solution & Example Answer:A uniform beam, AB, is held in a horizontal position by two vertical inelastic strings attached at the points C and D respectively - Leaving Cert Applied Maths - Question 7 - 2015

Step 1

Calculate the tension in each of the strings

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Answer

To find the tensions in the strings at points C and D, we apply the principle of moments about point P.

Total Weight Calculation: The total downward force can be calculated as: $T_1 + T_2 + 18 + 20 + 10 = 48$
Moments about Point P (Pivot):
- The moment due to T2 is calculated as: $T_2 \times 2 \text{ m} = 18 \times 0.5 + 20 \times 1 + 10 \times 1.5$
- Simplifying, we have: $T_2 \times 2 = 9 + 20 + 15 = 44$ Now solve for T2: $T_2 = 22 \text{ N}$
Finding T1: Now substitute T2 into the total weight equation: $T_1 + 22 + 18 + 10 = 48$ Thus: $T_1 = 26 \text{ N}$

Step 2

Find the coefficient of friction between the ladder and the ground

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Answer

Force Analysis: The forces acting on the ladder include its weight and the normal forces at the ground and the wall. Let:
- Weight of the ladder: $W = 160 \text{ N}$
- Normal force at the ground: $R_1$
- Normal force at the wall: $R_2$
Using Trigonometry: The ladder's configuration gives: $R_2 \times 8 \sin \theta = 160 \cos \theta$ From this, we derive: $R_2 = 80$
Setting Up for Friction: The vertical force balance gives: $R_1 = 160 \text{ N}$
Friction Coefficient Calculation: The frictional force at the ground provides: $\mu R_1 = \frac{R_2}{R_1}$ Hence: $\mu = \frac{R_2}{R_1} = \frac{80}{160} = 0.5 → 0.17$ So, the coefficient of friction between the ladder and the ground is approximately 0.17.

A uniform beam, AB, is held in a horizontal position by two vertical inelastic strings attached at the points C and D respectively - Leaving Cert Applied Maths - Question 7 - 2015

Worked Solution & Example Answer:A uniform beam, AB, is held in a horizontal position by two vertical inelastic strings attached at the points C and D respectively - Leaving Cert Applied Maths - Question 7 - 2015

Calculate the tension in each of the strings

Find the coefficient of friction between the ladder and the ground

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