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A uniform ladder, of weight 210 N, rests on rough horizontal ground and leans against a smooth vertical wall - Leaving Cert Applied Maths - Question 7 - 2017

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A uniform ladder, of weight 210 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 210 N, rests on rough horizontal ground and leans against a smooth vertical wall - Leaving Cert Applied Maths - Question 7 - 2017

Step 1

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

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Answer

To find the coefficient of friction (μ\mu) between the ladder and the ground, we start by analyzing the forces acting on the ladder.

Step 1: Analyze Forces

The forces acting on the ladder include:

  • Weight of the ladder (W) = 210 N acting downward at the center of gravity (midpoint).
  • Normal reaction (R) from the ground acting upward.
  • Frictional force (S) acting horizontally at the base in the opposite direction to potential slipping.

Step 2: Set up Equilibrium Equations

Since the ladder is in equilibrium:

  1. The sum of vertical forces must equal zero: RW=0R=W=210NR - W = 0 \quad \Rightarrow \quad R = W = 210 N
  2. The moment about the base (point where ladder meets ground) must also be zero. We can use the distances to write the moment equation: S×7=W×1S×7=210×1S \times 7 = W \times 1 \Rightarrow S \times 7 = 210 \times 1 S=2107=30N\Rightarrow S = \frac{210}{7} = 30 N

Step 3: Identify the Coefficient of Friction

Using the relation for friction: μS=Rμ=RS=30210=17\mu S = R \Rightarrow \mu = \frac{R}{S} = \frac{30}{210} = \frac{1}{7}

Thus, the coefficient of friction is μ=17\mu = \frac{1}{7}.

Step 2

Show on a diagram the forces acting on the particle.

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Answer

The diagram should include the following:

  • A downward force representing the weight of the particle, marked as 123 N.
  • Two tension forces (TT and SS) acting at angles to the horizontal, connecting to the points on the ceiling.
  • Indicate the angles of the strings with the horizontal as α and β respectively.

Step 3

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

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Answer

For the particle in equilibrium, the forces can be resolved as follows:

Horizontal Forces Equation:

Tcos(α)=Scos(β)T \cos(\alpha) = S \cos(\beta)

Vertical Forces Equation:

$$ T \sin(\alpha) + S \sin(\beta) = 123 $.

Step 4

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

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Answer

We use the equations established earlier:

Step 1: Substitute for SS

From the horizontal forces equation, we can express SS in terms of TT: S=Tcos(α)cos(β)S = T \frac{\cos(\alpha)}{\cos(\beta)} Substituting this into the vertical forces equation gives: Tsin(α)+(Tcos(α)cos(β))sin(β)=123T \sin(\alpha) + \left( T \frac{\cos(\alpha)}{\cos(\beta)} \right) \sin(\beta) = 123

Step 2: Solve for T

Now simplify and isolate TT: After some rearrangement, we will find T=27NT = 27 N.

Step 3: Find SS

Substituting the value of TT back into either equation will yield: S=120NS = 120 N

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