ch19 moments

19.1 the turning effect of a force

  • moment = F d , about a point where d is the perpendicular distance of the force from P’s line of action, (Nm)

    • force at centre of mass x distance from the centre to that SPECIFIC point.

  • centre of mass is where the object’s weight acts:

    • uniform rod: midpoint,

    • uniform lamina: intersection of diagonals.

  • don’t forget to state clockwise/anti-clockwise,

  • resultant moment about a point = sum of clockwise and anticlockwise separately, then the difference between the two sums in the direction of the larger moment.

  • ‘how far a point is from each force’

  • clockwise = negative values

  • uniform rod = mass distributed evenly so centre of mass at midpoint,

    • non-uniform rod = not necessarily centre of mass at the midpoint.

  • rigid body is in equilibrium if:

    • total moment about any point = 0 (choose the one with more unknowns),

    • resultant force = 0.

  • CHECK TO FIND THE CENTRE CHEEKY BASTARDS

19.2 equilibrium

  • to counterbalance an object’s weight to stay in equilibrium, there’ll be either:

    • smooth supports RA and RB with normal reaction forces,

    • OR light strings with tension TA and TB.

    • ++ no rotation needs to be added if in equilibrium.

  • equilibrium = ZERO resultant force (& about any point),

  • you have a choice in points, choose the one with the least number of forces as that will be = 0 and eliminated :o .

  • equilibrium : sum of clockwise = sum of anticlockwise, about the same point.

  • modelling assumptions:

    • plank is uniform,

    • mass is a particle,

    • ropes are light

  • if let’s say there’s A B C D, where B and C are strings, if A is about to tilt the particle, you can take P(TB) = 0 and ignore TC

    • if mass on D, you still consider it and ignore TC

  • the mass of the particle is not mg, everything else like a person standing on the particle is mg.