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Eunil Won Dept. of Physics, Korea University 1 Ch 03 Force Movement of massive object Force Source of the move Velocity, acceleration Eunil Won Dept. of Physics, Korea University 2 1 평(坪) = ? ~ ~ 3.305 2 m 2 m 1.82 m x 1.82 m Eunil Won Dept. of Physics, Korea University Newton’s First Law Law of Inertia (mass) Any body will remain at rest of in motion in a straight line with a constant velocity unless acted upon by an outside force If the net force is zero, there is a reference frame with zero acceleration 3 Eunil Won Dept. of Physics, Korea University 4 Newton’s Second Law Law of Motion If a force is applied, the state of motion is changed dv ⎤ ⎡ dp d = (mv) = m ⎥ = ma newton ⇒ [N = kg ⋅ m/s 2 ] F=⎢ dt ⎦ ⎣ dt dt Eunil Won Dept. of Physics, Korea University 5 Newton’s Third Law Action-Reaction Law Whenever one body exerts a force on a second body, the second body exerts a force back on the first that is equal in magnitude and opposite in direction FAB = −FBA Eunil Won Dept. of Physics, Korea University 6 Example 3.1 A 440-g can of food on a frictionless level surface is observed to accelerate at a rate of 1.5 m/sec2. What is the force on the can? Fnet , ext = ma = (0.440 kg)(1.5 m/s 2 ) = 0.66 kg ⋅ m/s 2 Fnet , ext = 0.66 N Eunil Won Dept. of Physics, Korea University 7 Example 3.2 An orderly The orderly exerts a backward force of 100 N on the floor. What acceleration is produced, assuming the friction in the wheels is negligible? (question on the system I) An orderly mo = 85kg Gurney mg = 20kg Patient m p = 50kg Force that the orderly exerts Fa = 100N Acceleration a= = Ft Fa = mt mo + mg + m p 100 = 0.645m/s 2 85 + 20 + 50 Eunil Won Dept. of Physics, Korea University 8 Example 3.3 Calculate the force the orderly exerts on the gurney in Example 3.2 (Question on system 2) An orderly mo = 85kg Gurney mg = 20kg Patient m p = 50kg Force that the orderly exerts Fa = 100N Force Fo = ma = (mg + m p )a = (20 + 50)(0.645) = 45.2kgm/s 2 = 45.2 N Eunil Won Dept. of Physics, Korea University 9 Weight and normal force weight F = ma w = mg Normal force Ft = 0 N −w=0 ⇒ N =w Eunil Won Dept. of Physics, Korea University 10 Newton’s Universal Law of Gravitation mM F=G 2 r : there is a force of attraction between any two masses that is proportional to the product of the masses and inversely proportional to the distance between their centers of mass mM w = mg ⇔ G 2 r GM g= 2 r Eunil Won Dept. of Physics, Korea University 11 Eunil Won Dept. of Physics, Korea University 12 Friction : proportional to the normal force Static Friction fs = µs N Kinetic Friction f (= f k ) = µ k N µs f ≤ fs µs : coefficient of static friction µk : coefficient of kinetic friction Eunil Won Dept. of Physics, Korea University 13 Friction Coefficients of friction Rubber on dry concrete Rubber on wet concrete Wood on Wood Waxed wood on wet snow Metal on wood Steel on steel (dry) Steel on steel (oiled) Teflon on steel Bone lubricated with synovial fluid (활액) Shoes on wood Shoes on ice Ice on ice Steel on ice Kinetic 0.7 0.5 0.3 0.1 0.3 0.3 0.03 0.04 Static 1.0 0.7 0.5 0.14 0.5 0.6 0.05 0.04 0.015 0.016 0.7 0.05 0.03 0.02 0.9 0.1 0.1 0.4 Eunil Won Dept. of Physics, Korea University 14 Example 3.4 A 70-kg cross-country skier is on level ground in wet snow wearing waxed wooden skis. (a) What is the maximum force that can be exerted on the skier in a horizontal direction without causing her to move? (b) Once the skier is moving, what horizontal force is necessary to keep her moving at a steady velocity? w = mg = (70 kg)(9.8 m/s 2 ) = 686 N f = µs N = (0.14)(686 N) = 96 N Fnet , ext = 0 ⇒ F − f = 0 F = f = µk N = (0.1)(686 N) = 68.6 N Eunil Won Dept. of Physics, Korea University 15 Tension A tension is any force carried by a flexible string, rope, cable, chain etc Eunil Won Dept. of Physics, Korea University 16 Example 3.5 A child and basket with a total mass of 10 kg are suspended from a scale by a cord. Calculate the tension in the cord. Fnet , ext = T − w = 0 T = w = mg = (10 kg)(9.8 m/s 2 ) = 98 N Eunil Won Dept. of Physics, Korea University Tensions of a patient 17 Eunil Won Dept. of Physics, Korea University 18 Vector 1. Vectors have direction & magnitude ρ {a, a}, a = a 2. Two vectors are identical if both direction and magnitude a=b are same 3. Negative vector. 4. Zero vector. 5. Unit vector. −a 0 eˆ ⇒ eˆ = 1 Eunil Won Dept. of Physics, Korea University 19 Vector analysis a+b = s a b a+b = b+a s a + (b + c) = (a + b) + c a + ( −a) = 0 a−b = d a+0=a −b d a Eunil Won Dept. of Physics, Korea University 20 Components of Vectors y a + b = (a x i + a y j) + (bx i + by j) a ay ŷ x̂ = (a x + bx )i + (a y + by ) j x ax xˆ = i = (1,0,0) yˆ = j = (0,1,0) a = ax + ay + az = a x xˆ + a y yˆ + a z zˆ a + b = s = sxi + s y j s x = a x + bx s y = a y + by a x = a cosθ , a y = a sin θ ⎛ ay ⎞ θ = tan ⎜⎜ ⎟⎟ ⎝ ax ⎠ −1 Eunil Won Dept. of Physics, Korea University 21 Example 3.6 Calculate the total force exerted on a gurney F1x = F1 cos 20° = 70(0.939) = 65.78 F1 y = F1 sin 20° = 70(0.342) = 23.94 F2 x = F2 cos15° = 60(0.966) = 57.95 F2 y = F2 sin 15° = 60(0.259) = 15.53 Ft = F1 + F2 F1 = F1x i + F1 y j Ftx = F1x + F2 x = 123.73 Fty = F2 y − F2 y = 8.41 2 2 F2 = F2 x i + F2 y j Ft = Ftx + Fty = 124.01 Ft = ( F1x + F2 x )i + ( F2 y + F2 y ) j ⎛ Fty ⎞ θ = tan ⎜ F ⎟ = 3.89° tx ⎠ ⎝ −1 Eunil Won Dept. of Physics, Korea University 22 Example 3.7/3.8 Calculate the acceleration (cart+stuff=25 kg, frictional force=2.0 N) Acceleration of the cart Fn = Fh − f = 17.3 − 2.0 = 15.3 Fn 15.3 = = 0.61m/s 2 a= 25 m Normal Force Fy = 0 N − w − Fv = 0 N = w + Fv = mg + Fv Ft = Fh + Fv Fv + N + mg = 0 = (25)(9.8) + 10 = 255 N Eunil Won Dept. of Physics, Korea University Torque and Rotation 23 Eunil Won Dept. of Physics, Korea University 24 Torque The effectiveness of a force in producing a rotation is called torque τ = rFt = rF sin φ = r⊥ F τ = r × F , τˆ //ω̂ Eunil Won Dept. of Physics, Korea University 25 Equilibrium Condition ∑ F = 0, ∑τ =0 Eunil Won Dept. of Physics, Korea University 26 Example 3.9 (a) Calculate the force that the biceps muscle exerts to hold the book (b) Compare the force exerted by the biceps muscle with the weight of the forearm and book. (clockwise) (counterclockwise) τ cw, net = τ ccw, net (15 cm)( wa ) + (40 cm)( wt ) = (4.0 cm) FB (15cm)(ma )( g ) + (40cm)(mt )( g ) 4.0cm (15)(2.5kg )(9.8m/s 2 ) + (40)(4.0kg )(9.8m/s 2 ) = 4.0 367.5 N + 1568 N = = 484 N 4.0 FB = w = wa + wt = ma g + mt g w = 63.7 N 484 N 〉〉 63.7 N Eunil Won Dept. of Physics, Korea University 27 Example 3.10 Calculate the force exerted by the extensor muscle in the upper leg when lifting a weight as shown below Note: l⊥ = 35cm l′⊥ = 2.0cm τ w, net = τ ccw, net l⊥T = l⊥′ FM ⇒ FM = (l⊥ / l⊥′ )T FM = (l⊥ / l′⊥ )mg = (35cm/2.0cm)(98 N) = 1715 N Eunil Won Dept. of Physics, Korea University 28 Rotational Motion and Centripetal Force v2 Fc = mar = m r Eunil Won Dept. of Physics, Korea University 29 v p = v px i + v py j = v cosθ i + v sin θ j v q = vqx i + vqy j = v cosθ i − v sin θ j pq r (2θ ) ∆t = ≈ v v ax = ay = vqx − v px ∆t vqy − v py ∆t a = −a y j = = vq cosθ − v p cosθ ∆t − vq sin θ − v p sin θ ( 2 rθ ) / v =0 v 2 ⎛ sin θ ⎞ =− ⎜ ⎟ r ⎝ θ ⎠ ⎛ v2 ⎞ sin θ v 2 a y = lim a y = ⎜⎜ ⎟⎟ lim = θ →0 r ⎝ r ⎠ θ →0 θ Eunil Won Dept. of Physics, Korea University 30 Example 3.11 An ultracentrifuge spins at 50,000 revolutions per minute and the material being centrifuged is 5.0 cm from the pivot point. Calculate the centripetal acceleration and express it as a multiple of the acceleration of gravity (in g’s) v= s 50000 × 2π (0.05 m) = = 2.62 ×10 2 m/s 60 s t v 2 (2.62 × 10 2 m/s) 2 ac = = = 1.37 ×106 m/s 2 r 0.05m ac 1.37 ×106 m/s 2 5 = = 1 . 40 × 10 g 9.8 m/s 2 ac = (1.40 ×105 ) g