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Transcript
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 ΣF = ma ΣFx = max ΣFy = may Example 2: wooden block F1 = 80N 30o 20 kg F2 = 60N Find the magnitude and Direction of the block’s accel. 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 F1 F2 x comp y comp Example 4: one force acting on an object a = 4.0 m/s2 straight down F1 = 147N a = 4.0 m/s2 Force F1 F2 What are the x and y components of the second force? x comp 147N F2x y comp 0N 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 A B FAB = -FBA 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 Contact Forces Push Pull Apparent Weight • 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 FN = mg + ma Apparent True weight weight • If acceleration is upward, apparent weight is greater than true weight Fy = +FN – mg = ma FN – mg = ma FN = mg + ma 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 Fy = +FN – mg = ma 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 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 ΣFy = FN – mg = ma FN – 735N = (75kg)(-9.8m/s2) FN = 0 Example 12: a neck in traction T F m1 Both act on 1st vertebrae Dr. wants 34N to be F What is 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 35o 35o T2 F T1 + T2 = R R = 2Tcos35o R is determined 2.2 kg mass F = 2T cos35o = 2mg cos35o 2.2 kg 2(2.2kg)(9.8m/s2) cos 35o = 35N 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