7. (a) One end of a uniform ladder, of weight W, rests against a smooth vertical wall, and the other end rests on rough horizontal ground. The coefficient of friction between the ladder and the ground is μ. The ladder makes an angle α with the horizontal and is in a vertical plane which is perpendicular to the wall. Show that a person of weight 3W can safely climb to the top of the ladder if $$μ > \frac{7}{8 \tan α}$$ (b) Two uniform smooth spheres each of weight W and radius 0.5 m, rest inside a hollow cylinder of diameter 1.6 m. The cylinder is fixed with its base horizontal. (i) Show on separate diagrams the forces acting on each sphere. (ii) Find, in terms of W, the reaction between the two spheres. (iii) Find, in terms of W, the reaction between the lower sphere and the base of the cylinder.

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7. (a) One end of a uniform ladder, of weight W, rests against a smooth vertical wall, and the other end rests on rough horizontal ground - Leaving Cert Applied Maths - Question 7 - 2010

Question 7

7.-(a)-One-end-of-a-uniform-ladder,-of-weight-W,-rests-against-a-smooth-vertical-wall,-and-the-other-end-rests-on-rough-horizontal-ground-Leaving Cert Applied Maths-Question 7-2010.png

7. (a) One end of a uniform ladder, of weight W, rests against a smooth vertical wall, and the other end rests on rough horizontal ground. The coefficient of frictio... show full transcript

Worked Solution & Example Answer:7. (a) One end of a uniform ladder, of weight W, rests against a smooth vertical wall, and the other end rests on rough horizontal ground - Leaving Cert Applied Maths - Question 7 - 2010

Step 1

Show that a person of weight 3W can safely climb to the top of the ladder if

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Answer

To demonstrate that a person weighing $3W$ can safely climb to the top of the ladder, we need to consider the forces and moments acting on the ladder.

Forces Acting on the Ladder

Weight of the Ladder (W): Acts vertically downwards at its center of gravity.
Weight of the Person (3W): Acts vertically downward at the point where the person is climbing.
Normal Force (R1): Acts vertically upward from the ground at the base of the ladder.
Frictional Force (μR1): Acts horizontally at the base opposing any slippage.
Normal Reaction (R2): Acts perpendicular to the wall.

Moments about Point C

Using the equations of equilibrium, setting moments about point C (the point where the ladder touches the ground):

Sum of moments due to forces: $R2 \sin(α) = W (\frac{l}{2} cos(α)) + 3W(\frac{l}{2} cos(α))$
where l is the length of the ladder.
Rearranging gives us: $R2 \tan(α) = \frac{7W}{2}$

Conclusion

Thus, with the balance of forces and ensuring safety, we arrive at the relationship:
$μ \geq \frac{7}{8 \tan α}$
indicating that this must hold for the person of weight $3W$ to climb safely.

Step 2

Show on separate diagrams the forces acting on each sphere.

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Answer

Sphere A

Weight (W): Acting downwards at the center of the sphere.
Normal Reaction from Cylinder Surface (R4): Acting vertically upward.
Normal Reaction from Sphere B (R2): Acting horizontally towards Sphere B.

Illustration:

Draw a circle representing Sphere A with arrows for W pointing down, R4 pointing up, and R2 pointing sideways.

Sphere B

Weight (W): Acting downwards at the center of the sphere.
Normal Reaction from Cylinder Surface (R3): Acting vertically upward.
Normal Reaction from Sphere A (R2): Acting horizontally towards Sphere A.

Illustration:

Draw a circle representing Sphere B with arrows for W pointing down, R3 pointing up, and R2 pointing sideways.

Step 3

Find, in terms of W, the reaction between the two spheres.

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Answer

Reaction Between the Two Spheres

To find the reaction between the two spheres (R2), we can use equilibrium conditions on either sphere.

From Sphere A:

Balancing vertical forces: $R4 - W = 0 \Rightarrow R4 = W$

From Sphere B:

Balancing horizontal forces: $R2 = \frac{5W}{4}$
since the setup requires understanding the geometry and forces acting on the spheres.

Step 4

Find, in terms of W, the reaction between the lower sphere and the base of the cylinder.

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Answer

Reaction Between Lower Sphere and Base of Cylinder

The reaction force (R3) at the base of the lower sphere can be derived by using equilibrium conditions again.

Balance Vertical Forces for Sphere B:
$R3 - W - R2 = 0$
Where R2 is the force from Sphere A.
Substituting R2 found above: $R3 = W + \frac{5W}{4} = \frac{9W}{4}$
This gives us the reaction force between the lower sphere and the base.

7. (a) One end of a uniform ladder, of weight W, rests against a smooth vertical wall, and the other end rests on rough horizontal ground - Leaving Cert Applied Maths - Question 7 - 2010

Worked Solution & Example Answer:7. (a) One end of a uniform ladder, of weight W, rests against a smooth vertical wall, and the other end rests on rough horizontal ground - Leaving Cert Applied Maths - Question 7 - 2010

Show that a person of weight 3W can safely climb to the top of the ladder if

Forces Acting on the Ladder

Moments about Point C

Conclusion

Show on separate diagrams the forces acting on each sphere.

Sphere A

Sphere B

Find, in terms of W, the reaction between the two spheres.

Reaction Between the Two Spheres

Find, in terms of W, the reaction between the lower sphere and the base of the cylinder.

Reaction Between Lower Sphere and Base of Cylinder

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