Forces and Dynamics Equilibrium of Forces

Equilibrium of Forces Translational Equilibrium If every part of a system moves in a straight line at a constant speed, we say it is in translational equilibrium. This includes being at rest. This means that using the graphical method of vector addition for the forces acting on the body, always produces a closed loop: For a body to be in translational equilibrium, the resultant forces in any two perpendicular directions must be zero

A skier moving at constant speed down a piste: E.g. 1 A skier moving at constant speed down a piste: i. Situation Diagram: ii. FBFD for skier: Gravitational force from Earth on skier Frictional force from Earth on skier Contact force from Earth on skier

Gravitational force from Earth on skier Frictional force from Earth on skier Contact force from Earth on skier

E. g. 1 A 50kg mass on a slope. Friction prevents it from moving E.g. 1 A 50kg mass on a slope. Friction prevents it from moving. Determine the frictional force. FBFD for m: N 50kg m 30° F 30° mg Equilibrium parallel to slope: mg sin30 – F = 0 0.5 x 50 x 10 – F = 0 F = 250N m 30°

E.g. 2 A lamp of mass 5kg hangs from two strings, each at an angle 50° to the vertical as shown. Calculate the tension in the strings. Considering vertical equilibrium: T cos50 + T cos50 – 49 = 0 0.64 T + 0.64 T = 49 1.28 T = 49 T = 49/1.28 = 38.3N F.B.F.D: 5kg 50° 50° 50° T T 49N Note: In this example the tension is the same in both strings, because they are at the same angle.

E.g. 3 A 10kg mass hangs over a pulley, connected to a stationary mass M on a 30° frictionless slope. If the system is in equilibrium determine m. 10kg m 30° FBFD for m: N 98N 30° mg Equilibrium parallel to slope: mg sin30 – 98 = 0 0.5 mg – 98 = 0 mg = 196 so m = 20kg Extension: Now determine the normal reaction force by considering equilibrium perpendicular to the slope. m 30°

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