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Published byAllan Chambers Modified over 6 years ago
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Monday Jan 18
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Answers q 1-4 pg 132 1a) 12N b) 3.0 m/s2 2) The reaction force acts on the child, not on the wagon itself, so there is still a net force on the wagon. 3 A) person pushes on ground; ground pushes on person B) snowball exerts a force on back; back exerts force on snowball C) ball exerts a force on glove; glove exerts a force on ball D) wind exerts a force on window; window exerts a force on wind 4) 1.6 m/s2 at an angle of 65 degrees north of east
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Define weight and explain how it differs from mass.
Weight: a measure of the gravitational force exerted on an object; its value can change with the location of the object in the universe. Mass: is a measure of the amount of matter in an object. It is an inherent property of an object. Define normal force and explain why friction is proportional to the normal force. Normal Force: the force that acts on a surface in a direction perpendicular to the surface. Both the normal force and friction are dependent on mass. Experimental findings have shown that the normal force and friction force are proportional. Define static friction, kinetic friction and the coefficient of friction. Static friction: force that resists the initiation of sliding motion between two surfaces that are in contact and at rest Kinetic friction: the force that opposes the movement of two surfaces that are in contact and in motion Coefficient of friction: the ratio of the magnitude of the force of friction to the magnitude of the normal force (between two objects in contact)
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Page 137 Q1-3
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On a piece of paper to hand in:
Draw a free body diagram of: A) a tv sitting at rest B) a 400 kg elephant sitting on a ramp elevated at 10 degrees from the horizon
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Example problem pg 138
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Chapter 4 Sample Problem Overcoming Friction
Section 4 Everyday Forces Chapter 4 Sample Problem Overcoming Friction A student attaches a rope to a 20.0 kg box of books.He pulls with a force of 90.0 N at an angle of 30.0° with the horizontal. The coefficient of kinetic friction between the box and the sidewalk is Find the acceleration of the box.
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Sample Problem, continued
Section 4 Everyday Forces Chapter 4 Sample Problem, continued 1. Define Given: m = 20.0 kg k = 0.500 Fapplied = 90.0 N at = 30.0° Unknown: a = ? Diagram:
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Sample Problem, continued
Section 4 Everyday Forces Chapter 4 Sample Problem, continued 2. Plan Choose a convenient coordinate system, and find the x and y components of all forces. The diagram on the right shows the most convenient coordinate system, because the only force to resolve into components is Fapplied. Fapplied,y = (90.0 N)(sin 30.0º) = 45.0 N (upward) Fapplied,x = (90.0 N)(cos 30.0º) = 77.9 N (to the right)
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Sample Problem, continued
Section 4 Everyday Forces Chapter 4 Sample Problem, continued Choose an equation or situation: A. Find the normal force, Fn, by applying the condition of equilibrium in the vertical direction: Fy = 0 B. Calculate the force of kinetic friction on the box: Fk = kFn C. Apply Newton’s second law along the horizontal direction to find the acceleration of the box: Fx = max
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Sample Problem, continued
Section 4 Everyday Forces Chapter 4 Sample Problem, continued 3. Calculate A. To apply the condition of equilibrium in the vertical direction, you need to account for all of the forces in the y direction: Fg, Fn, and Fapplied,y. You know Fapplied,y and can use the box’s mass to find Fg. Fapplied,y = 45.0 N Fg = (20.0 kg)(9.81 m/s2) = 196 N Next, apply the equilibrium condition, Fy = 0, and solve for Fn. Fy = Fn + Fapplied,y – Fg = 0 Fn N – 196 N = 0 Fn = –45.0 N N = 151 N Tip: Remember to pay attention to the direction of forces. In this step, Fg is subtracted from Fn and Fapplied,y because Fg is directed downward.
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Sample Problem, continued
Section 4 Everyday Forces Chapter 4 Sample Problem, continued B. Use the normal force to find the force of kinetic friction. Fk = kFn = (0.500)(151 N) = 75.5 N C. Use Newton’s second law to determine the horizontal acceleration. a = 0.12 m/s2 to the right
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Section 4 Everyday Forces
Chapter 4 Q1-4 page 139
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