2. (a) add components (b) use Pythagorean theorem for magnitude, inverse tan for direction (c) use 2nd law
8. (a) resolve force into x and y components, write all forces in vector notation, add components. (b) see 2 (b)
12. Use conversion of 4.45 N/lb; mass = wt./g
14. mass doesn't depend on location, weight does
17. Draw free body diagram for each mass; net force = 0 for each, Write equation for each situation and solve simultaneously.
18. In all cases, no acceleration so forces are equal.
39. (a) asled= F/msled (b) 3rd law situation (c) write euqations for sled position and girl's position using appropriate accelerations. Since positions are the same, set them equal and solve for t. Substitute value for t into girl's equation and find position.
40. Draw free body diagrams for each block (a) write 2nd law equation for each block in terms of a. Since a is same for both blocks, set equations equal and solve for action-reaction force. (b) Same method.
42. Case 1: a1 = F/m Case 2: a2= Fcosq/m
70. Forces are Fa from air, upward, and mg, downward. Assume up is positive. Let M = mass before ballast is thrown out and
m = mass of ballast. Write 2nd law for both
situations.
Fa - mg = -Ma
Fa - (M - m)g = (M - m)a
Solve simultaneously
73. Since pulley is fricitonless and massless, tension (T) is same in all parts of rope. Call upward positive so g = -9.8 m/s2. (a) 2nd law gives 2T - mg = ma. Solve for T. (b) Now 2t = m(g + a), with a = 1.30 m/s2. (c) Now the force for constant speed is the weight of the system, and for acceleration, T - mg = ma (d) Force on pulley by ceiling is always 2T (3rd law)