More Vector Examples Answers 1. A box is being pulled across the floor by a rope at an angle of 30o to the horizontal. The tension in the rope is 50 newtons. (a) Determine Fh , the horizontal component of the force. adj 50 N cos 30 = 30o hyp Fh Fh 50 N SOH-CAH-TOA (50 N) 0.866 = 50 N Fh = 43 N

More Vector Examples Answers 1. A box is being pulled across the floor by a rope at an angle of 30o to the horizontal. The tension in the rope is 50 newtons. (b) Determine Fv, the vertical component of the force on the rope. opp 50 N Fv sin 30 = 30o hyp 43 N Fv 50 N (50 N) 0.500 = SOH-CAH-TOA 50 N Fv = 25 N

2. A plane is flying with a speed of 150 m/s at a bearing of 120o.

2. A plane is flying with a speed of 150 m/s at a bearing of 120o. ve 30o 180o 150 m/s (a) Determine the eastward component of the velocity. adj cos 30 = hyp ve 150 m/s ve = 130 m/s (150 m/s) 0.866 = 150 m/s

2. A plane is flying with a speed of 150 m/s at a bearing of 120o. vs 180o 150 m/s (b) Determine the southward component of the velocity. opp sin 30 = hyp vs 150 m/s vs = 75 m/s (150 m/s) 0.500 = 150 m/s

2. A plane is flying with a speed of 150 m/s at a bearing of 120o. (c) How far east has the plane flown after 10 seconds? d = v t de = ve t = ( 130 m/s )( 10 s ) de = 1300 m

3. A rope is tied to a box on the floor and is held at angle of 40o to the horizontal. The box has a mass of 5.0 kg. A force of 60 newtons is exerted on the box via the rope. (a) Determine the horizontal and vertical components of the force. 60 N 40o 5.0 kg

3. A rope is tied to a box on the floor and is held at angle of 40o to the horizontal. The box has a mass of 5.0 kg. A force of 60 newtons is exerted on the box via the rope. (a) Determine the horizontal and vertical components of the force. 60 N Fv 40o 5.0 kg Fh Fh : FV : opp adj sin 40 = cos 40 = hyp hyp Fh 60 N Fv 60 N (60 N) 0.766 = (60 N) 0.643 = 60 N 60 N Fh = 46.0 N Fv = 38.6 N

3. (b) Calculate the normal force. 5.0 kg 46.0 N FN = ? Wt. = m g = ( 5.0 kg )( 9.8 m/s2 ) = 49 N

3. (b) Calculate the normal force. 5.0 kg 46.0 N FN = ? Wt. = 49 N Fup = Fdown FN + Fv = Wt. FN + 38.6 N = 49 N - 38.6 N - 38.6 N FN = 10.4 N

3. (c) If the coefficient of friction is 0 3. (c) If the coefficient of friction is 0.60, find the acceleration of the box. 60 N 40o Ff 5.0 kg 46.0 N a = ? μ = 0.60 10.4 N

3. (c) If the coefficient of friction is 0 3. (c) If the coefficient of friction is 0.60, find the acceleration of the box. 60 N 40o Ff 5.0 kg 46.0 N a = ? Fnet = ? μ = 0.60 10.4 N Fnet Fnet = m a a = m m m Fnet = Fh + Ff Ff = μ FN = ( 0.60 )( 10.4 N ) = 6.24 N

3. (c) If the coefficient of friction is 0 3. (c) If the coefficient of friction is 0.60, find the acceleration of the box. 60 N 40o 5.0 kg 46.0 N a = ? 6.3 N Fnet = ? μ = 0.60 10.4 N Fnet 39.76 N Fnet = m a a = = m 5.0 kg a = 8.0 m/s2 Fnet = Fh + Ff = ( + 46.0 N ) + ( - 6.24 N ) = 39.76 N