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A uniform ladder, of weight 150 N, rests on rough horizontal ground and leans against a smooth vertical wall - Leaving Cert Applied Maths - Question 7 - 2013
Question 7
A uniform ladder, of weight 150 N, rests on rough horizontal ground and leans against a smooth vertical wall. The foot of the ladder is 2 m from the wall and the to... show full transcript
Worked Solution & Example Answer:A uniform ladder, of weight 150 N, rests on rough horizontal ground and leans against a smooth vertical wall - Leaving Cert Applied Maths - Question 7 - 2013
Step 1
Find the coefficient of friction between the ladder and the ground.
Answer
In this scenario, we can analyze the forces acting on the ladder. The weight of the ladder (W) acts vertically downwards at its center of gravity, which is 3 m above the ground. The normal reaction force (R) acts vertically upwards. Applying the equilibrium equations:
-
The vertical forces:
R = W
Here, W = 150 N, so R = 150 N.
-
The horizontal forces:
The frictional force (S) acts horizontally towards the wall. Since the ladder is on the verge of slipping, the moments about the point where the ladder touches the ground can be considered:
S imes 6 = W imes 2
Substituting the known values gives:
S imes 6 = 150 imes 2
Therefore,
S = rac{150 imes 2}{6} = 50 N.
For static equilibrium:
S = μR
Substituting the values, we find:
50 = μ imes 150
μ = rac{50}{150} = rac{1}{3}.
Thus, the coefficient of friction between the ladder and the ground is ( μ = \frac{1}{3} ).
Step 2
Show on a diagram the forces acting on the particle.
Answer
To visualize the forces on the particle:
- Draw the particle at the intersection of two strings.
- Label the downward force due to weight (45 N).
- Label the tension in string 1 as T₁ and the tension in string 2 as T₂ at angles α and β respectively.
Indicating gravitational force downward and tensions at angles upwards reflects the system in equilibrium.
Step 3
Step 4
Solve these equations to find the tension in each of the strings.
Answer
From the equations:
- We know that: [ T_1 \cos(α) = T_2 \cos(β) ]
- The angles α and β can be determined using the lengths of the strings and the horizontal distance.
Let's consider using the sine laws from the vertical: [ T_1 \sin(α) + T_2 \sin(β) = 45 ] Rearranging and substituting from the first equation: After solving the system, we find: [ T_1 = 27 N ] [ T_2 = 36 N ]