A uniform ladder, of weight 200 N, rests on rough horizontal ground and leans against a smooth vertical wall. The foot of the ladder is 3 m from the wall and the top of the ladder is 5 m above the ground. The ladder is in equilibrium and is on the point of slipping. Find the coefficient of friction between the ladder and the ground. Two light inextensible strings are tied to a particle weighing 50 N. The other ends of the strings are tied to two points on a horizontal ceiling. The strings make angles $\alpha$ and $\beta$ with the ceiling, as shown in the diagram. $$\tan \alpha = \frac{4}{3}$$ and $$\tan \beta = \frac{3}{4}$$ (i) Show on a diagram the forces acting on the particle. (ii) Write down the two equations that arise from resolving the forces horizontally and vertically. (iii) Solve these equations to find the tension in each of the strings.

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A uniform ladder, of weight 200 N, rests on rough horizontal ground and leans against a smooth vertical wall - Leaving Cert Applied Maths - Question 7 - 2009

Question 7

A uniform ladder, of weight 200 N, rests on rough horizontal ground and leans against a smooth vertical wall. The foot of the ladder is 3 m from the wall and the to... show full transcript

Worked Solution & Example Answer:A uniform ladder, of weight 200 N, rests on rough horizontal ground and leans against a smooth vertical wall - Leaving Cert Applied Maths - Question 7 - 2009

Step 1

Find the coefficient of friction between the ladder and the ground.

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Answer

Identify the Forces Involved: The uniform ladder experiences gravitational force acting downwards (its weight, 200 N) and normal force from the ground acting upwards (R). The frictional force (f) opposes the motion at the base, while the horizontal force from the wall is negligible since there's no resultant horizontal force acting on the ladder.
Taking Moments About the Base of the Ladder: Since the ladder is in equilibrium, we can sum the moments about the base. The moment arm for the weight of the ladder (200 N) is the horizontal distance from the wall (3 m).

$R_{1}(5) = 200(1.5)$

Thus, calculating gives us:

$R_{1} = \frac{200(1.5)}{5} = 60 N$
Setting Up the Equation for Normal Force and Friction: From equilibrium in the vertical direction, we have:

$R = 200 N$

And since we sum up forces horizontally (friction), we have:

$\mu R = R_{1}$

Substituting known values:

$\mu(200) = 60$

Solving gives:

$\mu = \frac{60}{200} = \frac{3}{10}$

Step 2

Show on a diagram the forces acting on the particle.

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Answer

Draw the Particle and Forces: Start by sketching the particle at the center of the diagram. Draw the downward force of weight, 50 N, acting directly downwards.
Show the Strings and Angles: Indicate the two strings tied to the particle making angles $\alpha$ and $\beta$ with the ceiling. Label the tension in string 1 as $T_1$ and in string 2 as $T_2$ .
Illustrate the Forces: Draw $T_1$ angled upwards to the left at angle $\alpha$ and $T_2$ angled upwards to the right at angle $\beta$ . Include the horizontal and vertical components of both tensions:
- For $T_1$ : Horizontal component is $T_1 \cos(\alpha)$ ; Vertical component is $T_1 \sin(\alpha)$ .
- For $T_2$ : Horizontal component is $T_2 \cos(\beta)$ ; Vertical component is $T_2 \sin(\beta)$ .

Step 3

Write down the two equations that arise from resolving the forces horizontally and vertically.

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Answer

Horizontal Component Equation:

From the horizontal equilibrium, we have:

$T_1 \cos(\alpha) = T_2 \cos(\beta)$
Vertical Component Equation:

From the vertical equilibrium:

$T_1 \sin(\alpha) + T_2 \sin(\beta) = 50$

Step 4

Solve these equations to find the tension in each of the strings.

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Answer

Substituting Known Angles: Use the given values for the tangents:

$\tan(\alpha) = \frac{4}{3} \Rightarrow \sin(\alpha) = \frac{4}{5}, \cos(\alpha) = \frac{3}{5}$

$\tan(\beta) = \frac{3}{4} \Rightarrow \sin(\beta) = \frac{3}{5}, \cos(\beta) = \frac{4}{5}$
Use the Horizontal Equation:
$\Rightarrow T_1 \cdot \frac{3}{5} = T_2 \cdot \frac{4}{5}$$ Hence: $$T_1 = \frac{4}{3} T_2$$$
Substituting in the Vertical Equation:

Plug this into the vertical component equation:

$\frac{4}{3} T_2 \cdot \frac{4}{5} + T_2 \cdot \frac{3}{5} = 50$

Simplifying gives us:

$\Rightarrow \frac{16}{15} T_2 + \frac{3}{5} T_2 = 50$

Combining terms results in:
$\Rightarrow \frac{25}{15} T_2 = 50\n T_2 = 30 N$$$
Finding $T_1$ :

Now substitute back into $T_1 = \frac{4}{3} T_2$ :

$T_1 = \frac{4}{3} \cdot 30 = 40 N$

Thus, the tensions are: $T_1 = 40 N$ and $T_2 = 30 N$ .

A uniform ladder, of weight 200 N, rests on rough horizontal ground and leans against a smooth vertical wall - Leaving Cert Applied Maths - Question 7 - 2009

Worked Solution & Example Answer:A uniform ladder, of weight 200 N, rests on rough horizontal ground and leans against a smooth vertical wall - Leaving Cert Applied Maths - Question 7 - 2009

Find the coefficient of friction between the ladder and the ground.

Show on a diagram the forces acting on the particle.

Write down the two equations that arise from resolving the forces horizontally and vertically.

Solve these equations to find the tension in each of the strings.

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