4.1.3 Centres of Mass

Problem: Moments in 2 Dimensions

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Given:

A metal lever of mass 5 kg and length 1.5 m is attached by a smooth hinge to a vertical wall.
The lever is held at an angle of 30° to the vertical by a horizontal force of magnitude F N applied at the other end of the lever.
The lever is modeled as a uniform rod. Find: The value of F.

Resolve forces into components parallel and perpendicular to the given distances.

5g \sin(30^\circ) \times 0.75 \text{ (distance to the pivot)} - F \cos(30^\circ) \times 1.5 = 0

F \cos(30^\circ) \times 1.5 = 5g \sin(30^\circ) \times 0.75

F = \frac{5g \sin(30^\circ) \times 0.75}{\cos(30^\circ) \times 1.5}

Simplifying:

F = \frac{49.3 \text{ N}}{2 \times 0.866} \approx :success[14.15 \text{ N}]

5g \times 0.75 \times \sin(30^\circ) - F \times 1.5 \times \cos(30^\circ) = 0

Solve the resulting equation: The calculation is the same as in Method 1, leading to:

F = \frac{49.3 \text{ N}}{6} \approx :success[14.15 \text{ N}]

Result: The value of F is approximately 14.15 N.

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Draw a clear free body diagram: Identify all forces acting on the ladder, including:

Resolve forces and take moments: Resolve forces horizontally and vertically, applying equilibrium conditions ( ∑ Fx = 0 , ∑ Fy = 0). To solve for unknowns, take moments about a point, often the base of the ladder, where one or more forces pass through, eliminating them from the equation.
Use friction carefully: For the ladder to remain in equilibrium, friction at the base must be sufficient to prevent slipping. Use F_friction = μN, where N is the normal force, and check if the friction is enough by comparing it with the horizontal forces.