aob and cod are two uniform rods, each of weight W, freely hinged at o - Leaving Cert Applied Maths - Question 7 - 2007

The forces acting on the disc include:

• The weight (W) acting downwards, which equals (100g).
• The normal force (N) acting perpendicular to the surface from the kerb.
• The applied force (F) at an angle θ to the horizontal.

The diagram should illustrate:

• A circle representing the disc labeled with its radius.
• A vertical line representing the weight acting downward.
• A line perpendicular to the kerb representing the normal force, directed vertically upwards at the contact point.
• An angled line representing the applied force F.

To find the least value of the applied force F:

1. Resolve Forces:

   • Break down F into horizontal and vertical components: $F_{horiz} = F \cos \theta$ $F_{vert} = F \sin \theta$

2. Apply Equation of Equilibrium in Vertical Direction:

   • The vertical forces give: $R + F \sin \theta + N = 100g$
   • The normal force here can be found to be equal to 0: $N = 0$ so: $F \sin \theta = 100g$

3. Apply Equation of Equilibrium in Horizontal Direction:

   • The horizontal forces give: $R = \frac{3}{4} F$

4. Substituting and Solving:

   • Substituting R into the equations, we can find: $F \sin \frac{3}{4} + 0 = 100g$

5. Calculate the Value of F:

   • Finally through calculations, we determine: $F = 80 N \text{ or } 784 N$.

Thus the least value of F required to lift the disc is 80 N.