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Newton’s Law of Motion

Newton’s Law of Motion . Newton’s first law; every object continues in its state of rest, or of uniform velocity in a straight line, as long as no net force acts on it Newton’s second law; relation between acceleration and force Newton’s third law;

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Newton’s Law of Motion

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  1. Newton’s Law of Motion • Newton’s first law; every object continues in its state of rest, or of uniform velocity in a straight line, as long as no net force acts on it • Newton’s second law; relation between acceleration and force • Newton’s third law; whenever one object exerts a force on a second object, the second exerts an equal force in the opposite direction on the first

  2. Weight - the Force of Gravity and the Normal Force • Weight, FGis the force exerted on an object by gravity. • Normal force, FNis the force exerted perpendicular to a surface. * box mass = 10.0 kg

  3. Weight - the Force of Gravity and the Normal Force • Example: A box of mass m is placed on a smooth incline that makes an angle θ with the horizontal. (a) determine the normal force on the box. (b) determine the box’s acceleration. (c) evaluate for a mass m = 10 kg and an incline of θ = 30°.

  4. Using Newton’s Law - Friction • Friction ・　always present when two solid surfaces slide along each other ・　sliding friction is called kinetic friction Ffr = μkFN μk : coefficient of kinetic friction ・　static friction applies when two surfaces are at rest with respect to each other, Ffr≤ μsFN μs : coefficient of static friction

  5. Using Newton’s Law - Friction • Example : 10.0kg box rests on a horizontal floor. The coefficient of static friction is 0.40 and the coefficient of kinetic friction is 0.30. Determine the force of friction acting on the box if a horizontal external applied force is exerted on it of magnitude: (a) 0, (b) 10 N, (c) 20 N, (d) 38 N, and (e) 40 N.

  6. Using Newton’s Law - Friction • Example : Two boxes are connected by a cord running over a pulley. The coefficient of kinetic friction between box A and the table is 0.20. We ignore the mass of the cord and pulley and any friction in the pulley, which means we can assume that a force applied to one end of the cord will have the same magnitude at the other end. We wish to find the acceleration, a, of the system, which will have the same magnitude for both boxes assuming the cord doesn’t stretch. As box B moves down, box A moves to the right.

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