Lecture 4 equilibrium_of_forces
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1) Translational equilibrium occurs when every part of a system moves in a straight line at constant speed, including being at rest. 2) For a body to be in translational equilibrium, the resultant forces in any two perpendicular directions must be zero. 3) Examples are provided to demonstrate calculating forces to satisfy equilibrium conditions, such as a skier moving at constant speed, a mass on a slope, and masses connected by strings or over a pulley. The forces are summed using free body force diagrams.
Lecture 4 equilibrium_of_forces
- 1. Forces and Dynamics Equilibrium of Forces
- 2. 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
- 3. E.g. 1 A skier moving at constant speed down a piste: i. Situation Diagram: ii. FBFD for skier: Contact force from Earth on skier Gravitational force from Earth on skier Frictional force from Earth on skier
- 4. Contact force from Earth on skier Gravitational force from Earth on skier Frictional force from Earth on skier
- 5. E.g. 1 A 50kg mass on a slope. Friction prevents it from moving. Determine the frictional force. m 30° mg F FBFD for m: Equilibrium parallel to slope: mg sin30 – F = 0 0.5 x 50 x 10 – F = 0 F = 250N m 30° 30° N
- 6. 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 5kg 50° 50° 50° 50° 49N F.B.F.D: T T Note: In this example the tension is the same in both strings, because they are at the same angle.
- 7. 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° mg 98N FBFD for m: Equilibrium parallel to slope: mg sin30 – 98 = 0 0.5 mg – 98 = 0 mg = 196 so m = 20kg m 30° 30° Extension: Now determine the normal reaction force by considering equilibrium perpendicular to the slope. N