Newton s Laws III. 1. Frictional force and Normal Force The symbol for friction is f and since friction is a force, the unit will be N


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1 Newton s Laws III Level : Physics I Teacher : Kim i) Discover what factors that determines the friction acting between the object and the surface ii) Practice various cases of motion using ΣF = ma when friction is present 1. Frictional force and Normal Force The symbol for friction is f and since friction is a force, the unit will be N There are two factors that determines how much friction( f ) will act on a sliding object on a rough surface. The roughness of the surface(μ), and the normal force(f N) f = μ F N rough surface! where f is the frictional force, μ is the coefficient of friction, F N is the normal force i) Coefficient of Friction  Coefficient of friction is ratio of the force of friction between two bodies and the force pressing them together  The coefficient of friction depends on the materials used; for example, ice on steel has a low coefficient of friction, while rubber on pavement has a high coefficient of friction.  Coefficients of friction range from near zero to greater than one under good conditions, a tire on concrete may have a coefficient of friction of 1.7. ii) Static friction & Kinetic friction  Static friction is friction between two solid objects that are not moving relative to each other. For example, static friction can prevent an object from sliding down a sloped surface. The static friction force must be overcome by an applied force before an object can move.  Kinetic friction occurs when two objects are moving relative to each other and rub together (like a sled on the ground). Kinetic friction is usually less than static friction. Material Coefficient of Static Friction(μ s) Coefficient of Kinetic Friction(μ k) Rubber on Concrete Steel on Steel Wood on Wood Ice on Ice
2 F=ma, f = μ F N 2. Various Cases of Motion with Friction I. Horizontal force on a flat surface Q1) Suppose that a 12kg box is pulled by a 45N force parallel to a rough surface. The coefficient of kinetic friction μ k is 0.2. Find the acceleration of the box. Q2) Suppose that a 4kg box is pulled by a 12N force parallel to a rough surface. The coefficient of kinetic friction μ k is Find the acceleration of the box. a) 5.76m/s 2 b) 3.11m/s 2 c) 1.24m/s 2 d) 0.49m/s 2 Q3) How much horizontal force is needed to pull a block(m=20kg) so that the block will have an acceleration of 3m/s 2 on a rough surface? The coefficient of kinetic friction μ k is 0.4. a) 138.4N b) 116.2N c) 95.4N d) 59.8N Q4) A horizontal force of 30N is required to push an object of 10kg to have an acceleration of 1.6m/s 2. What is the coefficient of kinetic friction μ k? a) b) c) d) 0.143
3 F=ma, f = μ F N II. Inclined Plane with Friction Q5) How much force is needed to give a 5kg block an acceleration of 2m/s 2 up the 30º incline plane? The force is parallel to the incline. The coefficient of kinetic friction of the inclined plane is μ k =0.3. Solution is below a) 26.5N b) 31.5N c) 38.8N d) 47.2N *~Step by step solution~* i) Draw a freebody diagram to identify the forces acting on the block ii) Draw xy coordinate. The xaxis must be parallel to the direction of motion iii) If any force is not aligned x or y axis, divide that force into components iv) Write the equations using ΣF x=ma x and ΣF y=ma y y Fgsinθ fk FN θ Fgcosθ x ΣF x= F app  F gsinθ  f k = ma x, ΣF y= F N  F gcosθ = 0 Fg Since f k = μ k F N and F N = mgcosθ and F g =mg => F app  mgsinθ  μ k mgcosθ = ma x v) Solve for F app F app = mgsinθ + μ k mgcosθ +ma x ( F g = mg) = sin cos = 47.2N ΣF = ma, f = μf N
4 Q6) i) How much force is needed to move a 10kg block at constant speed up along 40º inclined plane if μ k is 0.3? a) 85.5N b) 78.8N c) 64.7N d) 55.1N ii) How much force is needed to give a 10kg block an acceleration of 3m/s 2 up the 40º incline if μ k is 0.3? a) 85.5N b) 98.8N c) 115.5N d) 155.1N Q7) A force of 50N is pushing a block of 5kg up an inclined plane of 30º above the horizontal. If μ k is 0.3, what is the acceleration of the block? a) 1.38m/s 2 b) 1.86m/s 2 c) 2.55m/s 2 d) 2.96m/s 2 ΣF = ma, f = μf N
5 Q8) An 8kg box is released on a 30º inclined plane and accelerates down the plane. The coefficient of kinetic friction μ k = i) Find the acceleration of the box. (*~the box is sliding due to gravity, so there is no applied force) a) 0.32m/s 2 b) 0.56m/s 2 c) 0.74/s 2 d) 0.92m/s 2 ii) Find the frictional force impeding its motion. a) 56.5N b) 43.6N c) 36.7N d) 23.8N iii) Does the mass of the object affect the acceleration of the box as it slides? a) Yeah~! b) No way~! c) Depends~! III. Applied force at an angle on a flat surface *Q9) Suppose that a 70kg box is pulled by a 400N force at an angle of 30 to the horizontal. The coefficient of kinetic friction μ k is 0.5. Find the acceleration of the box. ( Note : F N mg ) a) 1.48m/s 2 b) 3.22m/s 2 c) 5.41m/s 2 d) 7.12m/s 2 30 *~Step by step solution~*
6 step1) Draw an xy coordinate. It is a rule to draw the xaxis parallel to the direction of the motion! sinθ Y FN step2) Identify the forces by drawing a freebody diagram step3) Divide any forces into components if necessary X fk θ cosθ step4) Use Newton s 2 nd law ; ΣF x=ma x and ΣF y=ma y => ΣF x=f appcosθ f k = ma x & ΣF y=f N +F appsinθ F g = 0 Fg step5) Express normal force F N using ΣF y! => from ΣF y we get F N =mg F appsinθ (F g=mg) step6) Define frictional force => f k =µ k F N = µ k (mg F appsinθ) step7) Insert f k in ΣF x and find acceleration a x => ΣF x=f appcosθ µ k (mg F appsinθ) = ma x. Solving for a x= F appcosθ µ k (mg F app sinθ) m = 400cos ( sin30 ) =1.48m/s 2 70 F=ma, f = μ F N IV. Finding the minimum force needed to move an object from rest Q10) A block of mass 10kg is resting on flat surface. The coefficient of static friction μ s=0.5. i) What is the minimum horizontal force needed to move the object? a) 49N b) 58N c) 67N d) 76N ii) If the block is pulled at an angle of 30. What is the minimum force needed to move the object? Compare answer with above and see if a greater force is needed to move the object or less a) 35.2N b) 43.9N c) 55.1N d) 66.7N 30 F=ma, f = μ F N
7 Q11) A block of mass 20kg is resting on surface. The coefficient of static friction μ s=0.6. i) What is the minimum horizontal force needed to move the object? a) 63.8N b) 82.9N c) 98.5N d) 117.6N ii) If the block is pulled at an angle of 30 to the surface. What is the minimum force needed to move the object? a) 35.2N b) 63.8N c) N d) 135.5N 30 V. Finding the force needed to maintain constant speed Q12) An object of mass 20kg is given a horizontal push on a flat surface. If the coefficient of kinetic friction is μ k =0.05, then how much force is needed to maintain constant speed? a) 9.8N b) 12.3N c) 16.3N d) 17.1N Q13) An object of mass 5kg on a flat surface is pulled at an angle of 30, where the coefficient of kinetic friction μ k= How much force is needed to move the object at a constant speed? a) 4N b) 6N 30 c) 8N d) 10N
8 F=ma, f = μ F N VI. TwoBlock System Algebra Review  Find the value of x, y for the following equations i) 2x y = 2, y = 4 ii) x + y = 2, 2x y = 1 Tension Force Forces are often applied by means of cables or ropes that are used to pull an object, like a force T being applied to the right of a rope attached to a block. T Each particle in the rope, in turn, applies a force to its neighbor. As a result, the force is actually directly applied to the block. We always assume the rope is tight and strong so that there will be no compression or expansion on the rope We always assume the mass of the rope is very small compared to block so that we can ignore it. Q14) Without Friction A m 1=10kg mass on a horizontal frictionfree air track table is accelerated by a string attached to another m 2=10kg mass hanging vertically from a pulley. What is the acceleration of the system of both masses? What is the tension force(t)? Solution in the next page a) a=4.9m/s 2, T=98N b) a=4.9m/s 2, T=49N c) a=2.4m/s 2, T=98N d) a=2.4m/s 2, T=49N m1 m2
9 *~Step by step solution~* step1) Identify the forces by drawing a freebody diagram for both blocks separately step2) Draw an xy coordinate for both blocks. step3) Use Newton s 2 nd law for both blocks separately: ΣF=ma i) m 1 X: ΣF x=t = m 1a and Y : ΣF y=f N F g = 0 (since no vertical motion) FN T Fg ii) m 2 X: no horizontal force exists Y : ΣF y= F g T = m 2a where F g =m 2g T Fg step4) Combine the equations for both masses and solve for a and T T = m 1a m 2g T = m 2a Q15) With Friction A m 1=6kg block is on a horizontal surface. The coefficient of kinetic friction is µ k=0.22. The mass of the hanging block is m 2=3kg. Find the acceleration a of both blocks and the tension T. m1 m2 a) a=2.55m/s 2, T=18.2N b) a=1.83m/s 2, T=23.9N c) a=1.24m/s 2, T=28.1N d) a=0.76m/s 2, T=13.5N
10 F=ma, f = μ F N Q16) The coefficient of kinetic friction between block m 1 and the table is µ k=0.2. Also, m 1=25kg, m 2=15kg. How far will block m 2 drop in the first 3s after the system is released? (*~use d=½at 2 to find the distance fallen) m1 d=? m2 a) 11m b) 14m c) 18m d) 29m Q17) The coefficient of kinetic friction between block m 1 and the table is µ k=0.2. Also, m 1=25kg, m 2=15kg. A horizontal force of F app=250n is pulling on the m 1. What is the acceleration of the system? The system is moving to the left m1 m2 a) 4.55m/s 2 b) 3.49m/s 2 c) 2.14m/s 2 d) 1.35m/s 2
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