1. CHAPTER 13 Kinetics of a Particle:
Force and acceleration ( Newton’s method)

2. v = vo + act s = so + vot + act2 v2 = vo2 + 2ac(s-so) KINEMATICS
Rectilinear motion
Curvilinear motion
v = vo + act
s = so + vot + act2
v2 = vo2 + 2ac(s-so)
Normal & Tangential
Components
constant acceleration

3. CHAPTER 13

4. SF=0 SF=ma A particle originally at rest, will remain in rest
STATICS
Newton’s first law
A particle originally at rest, will remain in rest
SF=0
Newton’s second law
A particle acted upon by an unbalanced force,
experiences acceleration that has the same
direction as the force and a magnitude that is directly
proportional to the force
SF=ma
DYNAMICS

5. Kinetics diagram
Free body diagram

6. The crate has a mass of 50 kg. If the crate is subjected
Example 1
The crate has a mass of 50 kg. If the crate is subjected
to a 400[N] towing force as shown, determine the
velocity of the crate in 3[s] starting from rest.
ms= 0.5, mk= 0.3,

7. Equations of Motion :

8. Kinematics :
The acceleration is constant, P is constant

9. The crate has a mass of 80 kg. If the magnitude of P is
Example 2
The crate has a mass of 80 kg. If the magnitude of P is
increased until the crate begins to slide, determine the
crate’s initial acceleration. ms= 0.5, mk= 0.3,

10. Equations of equilibrium :
[solution]
80(9.81) = 784.8[N]
ms= 0.5
mk= 0.3 T
20o
Ff = 0.5N
: verge of slipping
: impending motion N
Equations of equilibrium :
SFx=0 ;
Tcos20o – 0.5N = ……..(i)
SFy=0 ;
N + Psin20o – = ……...(ii)
T = [N] , N = [N]

11. a [solution] 784.8 [N] 353.29 [N] Ff = 0.3N N Equations of Motion :
N – sin20o = 80(0)
N = [N]
SFy=may ;
SFx=max ;
353.29cos20o – 0.3(663.97) = 80a
a = 1.66 [m/s2]

13. Normal and Tangential Coordinates
Equations of Motion:
Normal and Tangential Coordinates o t
When a particle moves along a curved path, it may be more convenient to write the equation of motion in terms of normal and tangential coordinates.

14. Fn : centripertal force Fb : binormal force
SFbub o t
SFtut
SFnun
the force components acting on the particle
SFt = mat
SFn = man
SFb = 0
Ft : tangential force
Fn : centripertal force
Fb : binormal force

15. Example 1
The 3 kg disk is D is attached to the end of a cord as shown.
The other end of the cord is attached to a ball-and-socket joint
located at the center of the platform. If the platform is rotating rapidly,
and the disk is placed on it and released from rest as shown, determine
the time it takes for the disk to reach a speed great enough to break
the cord. The maximum tension the cord can sustain is 100 N. mk = 0.1

16. Frictional force : F = mkND
Direction : oppose the relative motion of the disk with
respect to the platform

17. Equations of Motion :
SFn=man ;
T = m(v2/r)
= 3v2
0.1ND = 3at
SFt=mat ;
SFb=0;
ND – = 0
ND = N
at = m/s2
vcr = 5.77 m/s
Setting T = 100 N to get critical speed vcr
Kinematics :
Since at is constant,
vcr = vo + att
5.77 = 0 + (0.981)t
T = 5.89 s

18. Example 2
At the instant q = 60o, the boys center of mass
is momentarily at rest.
Determine the speed and the tension in each of the
two supporting cords of the swing when q = 90o.
The boy has a weight of 300 N ( ≈ 30 kg).
Neglect his size and the mass of the seat and cords

19. q Free-body diagram Kinetic diagram n mat n t man 2T t W
At q = 60o, v = 0
At q = 90o, v = ?
T = ?
Free-body diagram
Kinetic diagram n
mat n t
man 2T q t W

20. q n n 2T man θ mat t W t Equations of Motion : an = v2/r = v2/3
SFn=man ;
2T – Wsinq = man
2T – 300sinq = 30(v2/3) ……(i)
Wcosq = mat
300cosq = 30at
10cosq = at ……(ii)
SFt=mat ;

21. 2T – 300sinq = 30(v2/3) ……(i)
10cosq = at ……(ii)
Kinematics : (to relate at and v)
vdv = at ds
vdv = 10cosq dq
v = 2.68 [m/s]
s = rq
ds = rdq
The speed of the boy at q = 90o
Solving, we get T = 186 [N]

22. Quiz 1
The 400 [kg] mine car is hoisted up the incline using the cable and motor M. For a short time, the force in the cable is F = 3200t 2 [N], where t is in seconds. If the car has an initial velocity v1=2 [m/s] at s=0 and t=0, determine the distance it moves up the plane when t=2 [s].
Ans = 5.43 m

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