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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.

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forces and dynamics

Forces and Dynamics

Equilibrium of Forces

slide2

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

slide3

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

slide4

Gravitational force from Earth on skier

Frictional force from Earth on skier

Contact force from Earth on skier

slide5

50kg

m

30°

E.g. 1A 50kg mass on a slope. Friction prevents it from moving. Determine the frictional force.

FBFD for m:

N

F

30°

mg

Equilibrium parallel to slope:

mg sin30 – F = 0

0.5 x 50 x 10 – F = 0

F = 250N

m

30°

slide6

50°

50°

5kg

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:

50°

50°

T

T

49N

Note:In this example the tension is the same in both strings, because they are at the same angle.

slide7

10kg

m

30°

E.g. 3A 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.

FBFD for m:

N

98N

30°

mg

Extension: Now determine the normal reaction force by considering equilibrium perpendicular to the slope.

Equilibrium parallel to slope:

mg sin30 – 98 = 0

0.5 mg – 98 = 0

mg = 196 so m = 20kg

m

30°