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Linear Motion Newton’s 2nd Law

Linear Motion Newton’s 2nd Law. Engineering 1h Dynamics. Velocity. Velocity is defined as the rate of change of distance with time. For a constant velocity, we can write:. where s is the position at time, t, and s 0 and t 0 are the starting position and time respectively.

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Linear Motion Newton’s 2nd Law

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  1. Linear Motion Newton’s 2nd Law Engineering 1h Dynamics

  2. Velocity Velocity is defined as the rate of change of distance with time. For a constant velocity, we can write: where s is the position at time, t, and s0 and t0 are the starting position and time respectively More generally, if the velocity is not constant, then for any instant,

  3. Acceleration Acceleration is defined as the rate of change of velocity with time. For a constant acceleration, we can write: If we start at time t0 = 0, then this can be re-written as: To find the distance travelled, we integrate the velocity:

  4. Acceleration Velocity is rate of change of distance, u is the starting velocity and doesn’t change, and we are assuming constant acceleration, so We can remove t if we know v

  5. Newton II The resultant force on a particle is equal to the rate of change of momentum of the particle Units: 1 Newton accelerates 1kg at 1ms-2. The form F=ma is only valid if the mass is constant.

  6. Example • A crate of mass 800kg is on a slipway inclined at 25o to the horizontal. Will the crate slip: • a) by itself (s=0.65) • b) when prodded (k=0.45) • Steps: • 1. draw the physical system • 2. draw the FBD • 3. Calculate R and hence Fmax • 4. Calculate F and compare Fmax

  7. Example - sketch and FBD sketch: F.B.D: Axes: F mg 25o R 25o mgcos25o mg Components of g: mgsin25o

  8. Example - solution  R = 7110N To find R: R - mgcos25o = 0 Therefore, for the static case, the maximum friction force available to stop the crate slipping is: Fmax = sR = 0.65 *7110 = 4620N The force needed to stop the crate slipping can be found from summing the forces up and down the slope: (zero, since the crate is not moving) mgsin25o-F = 0  F = 3310N Since Fmax is greater than the force required, F, the crate will not slip and so the ‘static’ assumption was correct

  9. Example - crate ‘nudged’ For this case, we assume that the crate has been nudged in some way and examine whether it will continue to accelerate, or come to a halt. As the crate is slipping, we use the coefficient of kinetic friction, k=0.45. F= kR = 0.45*7110 = 3200N This is less than the force required, so the crate will continue to slip.

  10. Acceleration of crate The crate will now move according to Newton II. mgsin25o-F = ma a = gsin25o - kR/m = 9.8 sin 25o - 0.45*7710/800 = 0.14ms-2 After being nudged, the crate will accelerate down the slope at 0.14ms-2.

  11. Systems of connected bodies Consider two blocks in contact with each other on a surface P A B What is the acceleration of the blocks? FBD P mAg mBg FA FB RB RA

  12. F F P (mA + mB) Acceleration of connected bodies Forces are in balance vertically RA = mAg RB = mBg FA=ARA FB=BRB friction forces  FA=A mAg FB=B mBg The body is moving horizontally, so Newton II applies P - FA - FB = (mA + mB)a  P - A mAg- B mBg = (mA + mB)a (A mA- B mB)g  a = - (mA + mB)

  13. F Forces between connected bodies To get the interactive force, treat as two separate systems P s s mAg mBg FA FB RA RB Vertical forces are as before. The horizontal forces are: P - FA - s = mA a s - FB = mB a adding these equations gives the same answer for the acceleration as before and s can be found from either equation s = mB a - FB

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