FORCES AND NEWTON’S LAWS OF MOTION

Newton’s First Law An object at rest will remain at rest and an object in motion will remain in motion unless acted upon by an outside force Net force is the vector sum of the forces acting on an object If Net Forces = Zero. Object is in Equilibrium

Inertia The natural tendency of an object to remain at rest or in motion at a constant speed along a straight line The mass of an object is a quantitative measure of inertia SI unit of Inertia and mass: kilogram (kg)

Newton’s Second Law of Motion If a net force is acting upon an object, the sum of the forces acting on the object is equal to the mass of the object times its acceleration ΣFnet = F1 + F2 + F3…= ma Fnet = ma

Example 1: A 900 kg car accelerates from rest to 25 m/s in 6 s. What is the net force acting on the car? SI unit of force = 1 Newton = 1kg∙m/s2

Find the magnitude and Direction of the block’s accel. ΣF = ma ΣFx = max ΣFy = may Example 2: wooden block Find the magnitude and Direction of the block’s accel. F1 = 80N 30o 20 kg F2 = 60N Force x comp y comp F1 -80N 0N F2 (cos 30o)(60N) (sin 30o)(60N) ΣFx= -28.04N ΣFy = -30N -28.04N = 20kg*ax -30N = 20kg*ay

Example 3: F1 = 28 N 25o F2 = 10 N Force x comp y comp F1 F2

Example 4: one force acting on an object a = 4.0 m/s2 straight down What are the x and y components of the second force? F1 = 147N a = 4.0 m/s2 Force x comp y comp F1 147N 0N F2 F2x F2y ΣFx = 147N + F2x ΣFy = 0 + F2y = may ax = 0 ΣFy = F2y = may ΣFx = 0 = 147N + F2x ΣFy = 63kg ∙ -4.0 m/s2 F2x = -147N F2y = -252N

Newton’s Third Law If Object A exerts a force on Object B, then Object B exerts a force on Object A. These forces are equal in magnitude but opposite in direction. FAB FBA A B FAB = -FBA

Example 5: a skater pushing against a wall A key to the correct application of the third law is that the forces are exerted on different objects. Make sure you don’t use them as if they were acting on the same object.

Identifying Forces Non-contact Forces Gravity Magnetism electric Push Pull

Apparent Weight ΣFy = +FN – mg = ma The force that the object exerts on the scale with which it is in contact Two forces acting W=mg Normal force (FN) ΣFy = +FN – mg = ma

Apparent Weight If acceleration is 0 m/s2, Fy = +FN – mg = ma If acceleration is upward, apparent weight is greater than true weight FN – mg = ma Apparent weight True

Apparent Weight If acceleration is downward, apparent weight is less than true weight Fy = +FN – mg = ma FN = -ma + mg If acceleration is equal to gravity (free fall), apparent weight is 0 FN – mg = -ma FN = 0

Example 11: 735N man in an elevator (75kg) Acceleration = 0 ΣFy = 0 ΣFy = FN – mg = ma 735N – (75kg)(9.8m/s2) = 0 735N = 735N Acceleration = 3.0 m/s2 upward ΣFy = FN – mg = (75kg)(3.0m/s2) FN -735N = 225N FN = 960N

FN – (75kg)(9.8m/s2) = (75kg)(-2.5m/s2) Acceleration = 2.5m/s2 downward ΣFy = FN – mg = ma FN – (75kg)(9.8m/s2) = (75kg)(-2.5m/s2) FN – 735N = -187.5N FN = 547.5N Acceleration = 9.8m/s2 FN – 735N = (75kg)(-9.8m/s2) FN = 0

Example 12: a neck in traction F Both act on 1st vertebrae Dr. wants 34N to be F What is m1? m1 ΣFx = F – T F = T T = mg F/g = m 34N / 9.8m/s2 = 3.5 kg

Example 13: a foot in traction ΣFy = T1 sin35 – T2 sin35 = 0 ΣFx = T1 cos35 + T2 cos35 - F = 0 T1 = T2 = T solve for F T1 + T2 = R R = 2Tcos35o R is determined 2.2 kg mass F = 2T cos35o = 2mg cos35o 2(2.2kg)(9.8m/s2) cos 35o = 35N T1 35o 35o T2 F 2.2 kg

Example 14: pulling a boat Find the sum of the two vectors FAx = FA cos45o = (40.0N)(0.707) = 28.3N FAy = FA sin45o = (40.0N)(0.707) = 28.3N FBx = FB cos37o = (30.0N)(0.799) =24.0N FBy = FB sin37o = -(30.0N)(0.602) = -18.1N FRx = FAx + FBx = 28.3N + 24.0N = 52.3N FRy = FAy + FBy = 28.3N -18.1N = 10.2N Then use Pythagorean theorem and tan-1