KINETICS of PARTICLESNewton’s 2nd Law &

The Equation of Motion

Lecture VI

aF m aF mOr

Newton’s 2nd Law &The Equation of Motion

Kinetics is the study of the relations between the unbalanced forces and the changes in motion that they produce.

Newton’s 2nd law states that the particle will accelerate when it is subjected to unbalanced forces. The acceleration of the particle is always in the direction of the applied forces.

Newton’s 2nd law is also known as the equation of motion. To solve the equation of motion, the choice of an appropriate coordinate systems

depends on the type of motion involved. Two types of problems are encountered when applying this equation:

The acceleration of the particle is either specified or can be determined directly from known kinematic conditions. Then, the corresponding forces, which are acting on the particle, will be determined by direct substitution.

The forces acting on the particle are specified, then the resulting motion will be determined. Note that, if the forces are constant, the acceleration is also constant and is easily found from the equation of motion. However, if the forces are functions of time, position, or velocity, the equation of motion becomes a differential equation which must be integrated to determine the velocity and displacement.

In general, there are three general approaches to solve the equation of motion: the direct application of Newton’s 2nd law, the use of the work & energy principles, and the impulse and momentum method.

Newton’s 2nd Law &The Equation of Motion

(Cont.)

Kinetic Diagram

Free-body Diagram

=ma

F2

F1

FR = F

P P

Note: The equation of motion has to be applied in such way that the measurements of acceleration are made from a Newtonian or inertial frame of reference. This coordinate does not rotate and is either fixed or translates in a given direction with a constant velocity (zero acceleration).

Newton’s 2nd Law &The Equation of Motion

(Cont.)

0

y

xx

F

maF

22

22

F

jiF

a

jia

yx

yx

yx

yx

yy

xx

FF

FF

aa

aa

maF

maF

Rectilinear Motion

Curvilinear Motion

Rectangular

Coordinates

n-t Coordinate

s

Polar Coordinate

s

22

22

F

eeF

a

eea

nt

nntt

nt

nntt

nn

tt

FF

FF

aa

aa

maF

maF

22

22

F

eeF

a

eea

FF

FF

aa

aa

maF

maF

r

rr

r

rr

rr

Newton’s 2nd Law &

The Equation of Motion Exercises

Exercise # 1

3/1: The 50-kg crate is projected along the floor with an initial speed of 7 m/s at x = 0. The coefficient of kinetic friction is 0.40. Calculate the time required for the crate to come to rest and the corresponding distance x traveled.

Exercise # 2

3/2: The 50-kg crate of Prob. 3/1 is now projected down an incline as shown with an initial speed of 7 m/s. Investigate the time t required for the crate to come to rest and the corresponding distance x traveled if (a) = 15° (b) = 30°.

Exercise # 3

3/4: During a brake test, the rear-engine car is stopped from an initial speed of 100 km/h in a distance of 50 m. If it is known that all four wheels contribute equally to the braking force, determine the braking force F at each wheel. Assume

a constant deceleration for the 1500-kg car.

Exercise # 4

3/6: The 300-Mg jet airliner has three engines, each of which produces a nearly constant thrust of 240 kN during the takeoff roll. Determine the length s of runway required if the takeoff speed is 220 km/h. Compute s first for an uphill takeoff direction from A to B and second for a downhill takeoff from B to A on the slightly

inclined runway. Neglect air and rolling resistance.

Exercise # 5

3/17: The coefficient of static friction between the flat bed of the truck and the crate it carries is 0.30. Determine the minimum stopping distance s which the truck can have from a speed of 70 km/h with constant deceleration if the crate is not to slip forward.

Problem 3/17

Exercise # 63/28: The system is released from rest with the cable taut .

For the friction coefficients s = 0.25 and k = 0.20, calculate the acceleration of each body and the tension T in the cable.

Neglect the small mass and fric tion of the pulleys. 20 kg

B

Problem 3/28

Exercise # 73/43: The sliders A and B are connected by a light rigid bar of length l = 0.5 m and move with negligible friction in the horizontal slots shown. For the position where xA = 0.4 m, the velocity of A is vA = 0.9 m/s to the right. Determine the acceleration of each slider and the force in the bar at this instant.

Problem 3/43

P = 40 N

Exercise # 8

3/51: If the 80-kg ski-jumper attains a speed of 25 m/s as he approaches the takeoff position, calculate the magnitude N of the normal force exerted by the snow on his skis just before he reaches A.

Exercise # 93/54: The hollow tube is pivoted about a horizontal axis through point O and is made to rotate in the vertical plane with a constant

counterclockwise angular velocity . = 3 rad/s. If a 0.l-kg particle is sliding in the tube toward O with a velocity of 1.2 m/s relative to the tube when the position = 30° is passed, calculate the magnitude N of the normal force exerted by the wall of the tube on the particle at this instant.

Problem 3/54

Exercise # 103/71: A small object A is held against the vertical side of the rotating cylindrical container of radius r by centrifugal action. If the coefficient of static friction between the object and the container is s, determine

the expression for the minimum rotational rate . = of the container which will keep the object from slipping down the vertical side.