Forces and Dynamics

1 / 13

# Forces and Dynamics - PowerPoint PPT Presentation

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.

I am the owner, or an agent authorized to act on behalf of the owner, of the copyrighted work described.

## PowerPoint Slideshow about 'Forces and Dynamics' - selina

Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author.While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server.

- - - - - - - - - - - - - - - - - - - - - - - - - - E N D - - - - - - - - - - - - - - - - - - - - - - - - - -
Presentation Transcript

### 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

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

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°

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.

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°