Application of Newton’s First Law of Motion (Examples) (Problems taken from Young & Freedman, University Physics, 13 th ed.) A. Object at Rest Example 6. A large wrecking ball is held in place by two light steel cables as shown in Fig. 2. If the mass m of the ball is 4090 kg, what are the (a) tension TB in the cable that makes an angle of 40 o with the vertical and (b) tension TA in the horizontal cable? Solution: Figure 1.3. Example 6 A. Set-up: Draw FBD for m. TBy TB 40o TA TBx β Fg B. Execute: for horizontal forces or vertical forces ∑ πΉπ₯ = 0 ∑ πΉπ¦ = 0 ππ΅π₯ − ππ΄ = 0 ππ΅π¦ − πΉπ = 0 where πΉπ = ππ and ππ΅π¦ = ππ΅ πππ 40π where ππ΅π₯ = ππ΅ π ππ40π ππ΅ π ππ40π − ππ΄ = 0 ππ΅ π ππ40π = ππ΄ Equate (1) and (2): ππ΅ πππ 40π − ππ = 0 ππ΅ πππ 40π = ππ (1) ππ΅ π ππ40π ππ΄ = ππ΅ πππ 40π ππ ππ΄ = πππ‘ππ40π = (4090ππ) (9.8 ππ From (2) π ) (π‘ππ40π ) = π. ππππππ π΅ π 2 π (4090ππ) (9.8 ππ 2 ) ππ π ππ΅ = = = π. ππππππ π΅ πππ 40π πππ 40π (2) Example 7. A 1130-kg car is held in place by a light cable on a very smooth frictionless ramp, as shown in Fig. 3. The cable makes an angle of 31.0o above the surface of the ramp, and the ramp itself rises at 25 o above the horizontal. (a) Draw FBD for the car. (b) Find the tension in the cable. (c) How hard does the surface of the ramp push on the car? Figure 1.4. Example 7 car held in place Solution: (a) Set-up: Draw FBD for the car. T – tension in cable FN – normal force T FN Tsin31o β Tcos31o mgsin25o mgcos25o mg Execute: (b) Net force on car (parallel to the inclined plane) ∑ πΉππππππππ = 0 ππππ 31π − πππ ππ25π = 0 πππ ππ25π πππ 31π π (1130ππ) (9.8ππ 2 ) π ππ25π π π= πππ 31π π= π» = πππππ΅ (c) Net force on car (perpindicular to the inclined plane) ∑ πΉππππππππππ’πππ = 0 πΉπ + ππ ππ 31π − πππππ 25π = 0 πΉπ = πππππ 25π − ππ ππ 31π = 0 ππ΅ = πππππ΅ FN refers to the force the surface of the ramp exerts on the car. B. Objects Moving with Constant Velocity Example 8. A block with mass m 1 is placed on an inclined plane with slope angle πΌ and is connected to a second hanging block with mass m 2 by a cord passing over a small, frictionless pulley. The coefficient of static friction is ππ and the coefficient of kinetic friction between m 1 and inclined plane is ππ . (a) Find the mass m2 for which m1 moves up the plane at Figure 5. Example 8 constant speed once it is set in motion. (b) Find the mass m2 for which m1 moves down the plane at constant speed once it is set in motion. Solution: Set-up: Draw FBD for each block. a) m1 moves up the plane at constant speed FN T T fk β mgsinπΌ mgcos25o β m1g m2g m1g Block m1 Block m2 where T – tension in cord FN – normal force πΉπ = π1 ππππ 25π ππ = ππ πΉπ = π1 ππππ 25π Execute: Execute: Net force on m2 (parallel to the inclined plane) Net force on m2 ∑ πΉππππππππ = 0 ∑ πΉππππππππ = 0 π − π1 ππ πππΌ − ππ = 0 π − π2 π = 0 π − π1 ππ πππΌ − ππ π1 ππππ 25π = 0 π = π1 ππ πππΌ + ππ π1 ππππ 25π = 0 Equate (1) and (2): (1) π = π2 π (2) π2 π = π1 ππ πππΌ + ππ π1 ππππ 25π π2 = π1 π πππΌ + ππ π1 πππ 25π = π1 (π πππΌ + ππ πππ 25π ) (b) m1 moves down the plane at constant speed once it is set in motion. FBD for m1: FN T fk mgcos25o mgsinπΌ m1g m1g ∑ πΉππππππππ = 0 π − π1 ππ πππΌ + ππ = 0 π − π1 ππ πππΌ + ππ π1 ππππ 25π = 0 π = π1 ππ πππΌ − ππ π1 ππππ 25π = 0 (3) Equate (2) and (3) π2 π = π1 ππ πππΌ − ππ π1 ππππ 25π π2 = π1 π πππΌ − ππ π1 πππ 25π = π1 (π πππΌ − ππ πππ 25π )
