Newton’s Laws Applications. 2 nd Law Procedure 1. Draw a force or free body diagram. 2. Set up ΣF = ma equations for each dimension. 3. Use kinematics.

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Newton’s Laws Applications

2 nd Law Procedure 1. Draw a force or free body diagram. 2. Set up ΣF = ma equations for each dimension. 3. Use kinematics if necessary. 4. Substitute known quantities. 5. Calculate the unknown quantities.

A Mass m hangs by a cable from the bottom of an elevator. What is the tension in the cable when the elevator is stopped at a floor? A. g B. mg C. Greater in magnitude than mg D. Less magnitude than mg E. 0 m

Tension (elevator) The sum of the forces must be zero or the mass would accelerate! M T MgMg

A Mass m hangs by a cable from the bottom of an elevator. What is the tension in the cable when the elevator accelerates upward? A. g B. mg C. Greater in magnitude than mg D. Less magnitude than mg E. 0 m

Tension (elevator) T must be greater than mg in order to result in an upward acceleration! M T MgMg

A Mass m hangs by a cable from the bottom of an elevator. What is the tension in the cable when the elevator is moving at constant speed between floors? A. g B. mg C. Greater in magnitude than mg D. Less magnitude than mg E. 0 m

Tension (elevator) The sum of the forces must be zero because mass is not accelerating! M T MgMg

A Mass m hangs by a cable from the bottom of an elevator. What is the tension in the cable when the elevator is slowing down at the top floor? A. g B. mg C. Greater in magnitude than mg D. Less magnitude than mg E. 0 m

Tension (elevator) Net force must point down in order to accelerate down M T MgMg

A Mass m hangs by a cable from the bottom of an elevator. What will the tension in the cable be if it comes dettached from the elevator? A. g B. mg C. Greater in magnitude than mg D. Less magnitude than mg E. 0 m

Tension (elevator) No opposing force! (weightless???) M MgMg

“Magic” pulleys simply bend the coordinate system. Pulley problems m1m1 m2m2

-x x “Magic” pulleys simply bend the coordinate system. Pulley problems m1m1 m2m2 m1gm1g N T T m2gm2g  F = ma m2gm2g = (m 1 +m 2 )a