Magnetic force #2 - Questions and Answers

Question 1 Two adults and a child want to push a wheeled cart in the direction marked x in Fig. 4.29. The two adults push with horizontal forces 𝐹 1 F 1 ​ and 𝐹 2 F 2 ​ as shown in the figure. (a) Find the magnitude and direction of the smallest force that the child should exert. (b) The cart accelerates at 2.0   m/s 2 2.0m/s 2 in the + 𝑥 +x-direction. What is the weight of the cart? Answer 1 (a) Resultant forces along y-direction: The resultant must have no y-component, so the child must push with a force: 140 sin ⁡ ( 3 0 ∘ ) − 100 sin ⁡ ( 6 0 ∘ ) = 16.6   N . 140sin(30 ∘ )−100sin(60 ∘ )=16.6N. For the child to exert the smallest possible force, that force will have no x-component, so the smallest possible force has a magnitude of 16.6   N 16.6N at an angle of 27 0 ∘ 270 ∘ or 9 0 ∘ 90 ∘ clockwise from the + 𝑥 +x-direction. (b) Resultant forces along x-direction: 𝑚 = Σ 𝐹 𝑎 m= a ΣF ​ 𝑚 = 140 cos ⁡ ( 3 0 ∘ ) + 100 cos ⁡ ( 6 0 ∘ ) 2 = 85.6   kg . m= 2 140cos(30 ∘ )+100cos(60 ∘ ) ​ =85.6kg. 𝑤 = 𝑚 𝑔 = ( 85.6 ) ( 9.8 ) = 840   N . w=mg=(85.6)(9.8)=840N. Answer 2 (a) Forces in the 𝑦 y-direction: 𝐹 1 𝑦 = 100 sin ⁡ ( 6 0 ∘ ) = 100 ⋅ 3 2 = 86.6   N . F 1y ​ =100sin(60 ∘ )=100⋅ 2 3 ​ ​ =86.6N. 𝐹 2 𝑦 = − 140 sin ⁡ ( 3 0 ∘ ) = − 140 ⋅ 1 2 = − 70   N . F 2y ​ =−140sin(30 ∘ )=−140⋅ 2 1 ​ =−70N. So, the smallest force that the child should exert: 𝐹 𝑐 = 86.6 − 70 = 16.6   N at  9 0 ∘ ( clockwise from  − 𝑦 -direction ) . F c ​ =86.6−70=16.6N at 90 ∘ (clockwise from −y-direction). (b) Resultant forces along 𝑥 x-direction: 𝐹 𝑥 = 𝐹 1 cos ⁡ ( 6 0 ∘ ) + 𝐹 2 cos ⁡ ( 3 0 ∘ ) + 𝐹 𝑐 . F x ​ =F 1 ​ cos(60 ∘ )+F 2 ​ cos(30 ∘ )+F c ​ . 𝐹 𝑥 = 100 cos ⁡ ( 6 0 ∘ ) + 140 cos ⁡ ( 3 0 ∘ ) = 171.24   N . F x ​ =100cos(60 ∘ )+140cos(30 ∘ )=171.24N. Weight of the cart: 𝑚 = 𝐹 𝑥 𝑎 = 171.24 2 = 85.62   kg . m= a F x ​ ​ = 2 171.24 ​ =85.62kg. Question 2 A stream of water strikes a stationary turbine blade horizontally, as the drawing illustrates. The incident water stream has a velocity of + 18.0   m/s +18.0m/s, while the exiting water stream has a velocity of − 18.0   m/s −18.0m/s. The mass of water per second that strikes the blade is 17.0   kg/s 17.0kg/s. Find the magnitude of the average force exerted on the water by the blade. Answer: This is essentially a momentum/impulse problem. Using Newton's second law: 𝐹 = Δ 𝑝 Δ 𝑡 = 𝑚 Δ 𝑣 Δ 𝑡 . F= Δt Δp ​ = Δt mΔv ​ . Here: Δ 𝑝 = 𝑚 ( 𝑣 𝑓 − 𝑣 𝑖 ) = 17.0 ⋅ ( − 18.0 − 18.0 ) . Δp=m(v f ​ −v i ​ )=17.0⋅(−18.0−18.0). Δ 𝑝 = 17.0 ⋅ ( − 36.0 ) = − 612   kg ⋅ m/s . Δp=17.0⋅(−36.0)=−612kg⋅m/s. 𝐹 = − 612   N . F=−612N. The question asks for magnitude: 𝐹 = 612   N . F=612N. Question 3 Boxes A and B are in contact on a horizontal, frictionless surface. Box A has mass 20.0   kg 20.0kg and Box B has mass 5.0   kg 5.0kg. A horizontal force of 250   N 250N is exerted on Box A. What is the magnitude of the force that Box A exerts on Box B? Answer 1 Acceleration of the system: 𝑎 𝑥 = 𝐹 𝑚 1 + 𝑚 2 = 250 20 + 5 = 250 25 = 10   m/s 2 . a x ​ = m 1 ​ +m 2 ​ F ​ = 20+5 250 ​ = 25 250 ​ =10m/s 2 . Force exerted by A on B: 𝐹 𝐵 = 𝑚 𝐵 𝑎 𝑥 = 5 ⋅ 10 = 50   N . F B ​ =m B ​ a x ​ =5⋅10=50N. Answer 2 Acceleration of the system: 𝑎 𝑥 = 𝐹 𝑚 1 + 𝑚 2 = 250 25 = 10   m/s 2 . a x ​ = m 1 ​ +m 2 ​ F ​ = 25 250 ​ =10m/s 2 . Force exerted by A on B: 𝐹 𝐵 = 𝑚 𝐵 𝑎 𝑥 = 5 ⋅ 10 = 50   N . F B ​ =m B ​ a x ​ =5⋅10=50N. Answer 3 Step 1: Denote the force Box A exerts on Box B as 𝐹 𝐴 𝐵 F AB ​ . The system moves with the same acceleration: 𝑎 = 𝐹 𝑚 1 + 𝑚 2 = 250 25 = 10   m/s 2 . a= m 1 ​ +m 2 ​ F ​ = 25 250 ​ =10m/s 2 . Step 2: Apply Newton's second law for Box B: 𝐹 𝐴 𝐵 = 𝑚 𝐵 𝑎 = 5 ⋅ 10 = 50   N . F AB ​ =m B ​ a=5⋅10=50N. Answer 4 Using Newton’s third law: 𝐹 𝐴 𝐵 = − 𝐹 𝐵 𝐴 . F AB ​ =−F BA ​ . Here, Box B exerts − 50   N −50N on Box A.