Magnetic force #1 - Questions and Answers

Question 1 Two snowcats tow a housing unit to a new location at McMurdo Base, Antarctica, as shown in the figure. The sum of the forces 𝐹 𝐴 F A ​ and 𝐹 𝐵 F B ​ exerted on the unit by the horizontal cables is parallel to the line 𝐿 L, and 𝐹 𝐴 = 4200   N F A ​ =4200N. Determine 𝐹 𝐵 F B ​ . Determine the magnitude of 𝐹 𝐴 + 𝐹 𝐵 F A ​ +F B ​ . Answer: Sum of forces is along line 𝐿 L: The horizontal components balance: 𝐹 𝐴 sin ⁡ ( 3 0 ∘ ) = 𝐹 𝐵 sin ⁡ ( 5 0 ∘ ) F A ​ sin(30 ∘ )=F B ​ sin(50 ∘ ) 𝐹 𝐵 sin ⁡ ( 5 0 ∘ ) = 4200 sin ⁡ ( 3 0 ∘ ) F B ​ sin(50 ∘ )=4200sin(30 ∘ ) 𝐹 𝐵 = 4200 sin ⁡ ( 3 0 ∘ ) sin ⁡ ( 5 0 ∘ ) = 6434.773232   N F B ​ = sin(50 ∘ ) 4200sin(30 ∘ ) ​ =6434.773232N The net force 𝐹 𝐴 + 𝐹 𝐵 F A ​ +F B ​ is: 𝐹 𝐴 + 𝐹 𝐵 = 𝐹 𝐴 cos ⁡ ( 3 0 ∘ ) + 𝐹 𝐵 cos ⁡ ( 5 0 ∘ ) F A ​ +F B ​ =F A ​ cos(30 ∘ )+F B ​ cos(50 ∘ ) = 4200 cos ⁡ ( 3 0 ∘ ) + 6434.773232 cos ⁡ ( 5 0 ∘ ) =4200cos(30 ∘ )+6434.773232cos(50 ∘ ) = 8272.3851253   N =8272.3851253N Question 2 A potter's wheel having a radius of 0.50   m 0.50m and a moment of inertia of 12   kg ⋅ m 2 12kg⋅m 2 is rotating freely at 50   rev/min 50rev/min. The potter can stop the wheel in 6.0   s 6.0s by pressing a wet rag against the rim and exerting a radially inward force of 70   N 70N. Find the effective coefficient of kinetic friction between the wheel and the wet rag. Answer: The angular acceleration is: 𝛼 = 𝜔 𝑓 − 𝜔 𝑖 Δ 𝑡 = 0 − 𝜔 𝑖 6.0   s α= Δt ω f ​ −ω i ​ ​ = 6.0s 0−ω i ​ ​ Convert 𝜔 𝑖 ω i ​ to rad/s: 𝜔 𝑖 = 50   rev/min × 2 𝜋   rad 60 = 50 ⋅ 2 𝜋 60 = 5 𝜋 3   rad/s ω i ​ = 60 50rev/min×2πrad ​ = 60 50⋅2π ​ = 3 5π ​ rad/s 𝛼 = 0 − 5 𝜋 3 6 = − 5 𝜋 18   rad/s 2 α= 6 0− 3 5π ​ ​ =− 18 5π ​ rad/s 2 The torque is: 𝜏 = 𝐼 𝛼 = 12 ⋅ − 5 𝜋 18 = − 60 𝜋 18 = − 20.94   N ⋅ m τ=Iα=12⋅− 18 5π ​ =− 18 60π ​ =−20.94N⋅m The friction force is: 𝜏 = 𝑟 𝑓 𝑟    ⟹    𝑓 𝑟 = 𝜏 𝑟 = − 20.94 0.50 = − 41.88   N . τ=rf r ​ ⟹f r ​ = r τ ​ = 0.50 −20.94 ​ =−41.88N. The coefficient of kinetic friction is: 𝜇 𝑘 = 𝑓 𝑟 𝐹 𝑟 = 41.88 70 ≈ 0.30. μ k ​ = F r ​ f r ​ ​ = 70 41.88 ​ ≈0.30. Question 3 Resolve the force 𝐹 2 F 2 ​ into components acting along the 𝑢 u- and 𝑣 v-axes and determine the magnitudes of the components. Answer 1 Using sine rule: 𝐹 2 sin ⁡ ( 7 0 ∘ ) = 𝐹 2 𝑢 sin ⁡ ( 4 5 ∘ ) = 𝐹 2 𝑣 sin ⁡ ( 6 5 ∘ ) . sin(70 ∘ ) F 2 ​ ​ = sin(45 ∘ ) F 2u ​ ​ = sin(65 ∘ ) F 2v ​ ​ . Calculate the force component of 𝐹 2 F 2 ​ in 𝑢 u-direction: 𝐹 2 𝑢 = 𝐹 2 ⋅ sin ⁡ ( 4 5 ∘ ) sin ⁡ ( 7 0 ∘ ) = 500 ⋅ sin ⁡ ( 4 5 ∘ ) sin ⁡ ( 7 0 ∘ ) = 376.24   N . F 2u ​ =F 2 ​ ⋅ sin(70 ∘ ) sin(45 ∘ ) ​ =500⋅ sin(70 ∘ ) sin(45 ∘ ) ​ =376.24N. Calculate the force component of 𝐹 2 F 2 ​ in 𝑣 v-direction: 𝐹 2 𝑣 = 𝐹 2 ⋅ sin ⁡ ( 6 5 ∘ ) sin ⁡ ( 7 0 ∘ ) = 500 ⋅ sin ⁡ ( 6 5 ∘ ) sin ⁡ ( 7 0 ∘ ) = 482.24   N . F 2v ​ =F 2 ​ ⋅ sin(70 ∘ ) sin(65 ∘ ) ​ =500⋅ sin(70 ∘ ) sin(65 ∘ ) ​ =482.24N. The magnitude of 𝐹 2 F 2 ​ is: 𝐹 2 = 𝐹 2 𝑢 2 + 𝐹 2 𝑣 2 = ( 376.24 ) 2 + ( 482.24 ) 2 = 611.65   N . F 2 ​ = F 2u 2 ​ +F 2v 2 ​ ​ = (376.24) 2 +(482.24) 2 ​ =611.65N. Answer 2 Using sine rule: 𝐹 𝑢 sin ⁡ ( 4 5 ∘ ) = 𝐹 𝑧 sin ⁡ ( 3 0 ∘ ) = 𝐹 𝑣 sin ⁡ ( 7 5 ∘ ) . sin(45 ∘ ) F u ​ ​ = sin(30 ∘ ) F z ​ ​ = sin(75 ∘ ) F v ​ ​ . Calculate 𝐹 𝑢 F u ​ and 𝐹 𝑣 F v ​ : 𝐹 𝑢 = 𝐹 𝑧 sin ⁡ ( 4 5 ∘ ) sin ⁡ ( 7 5 ∘ ) = 6   kN , F u ​ = sin(75 ∘ ) F z ​ sin(45 ∘ ) ​ =6kN, 𝐹 𝑣 = 𝐹 𝑧 sin ⁡ ( 3 0 ∘ ) sin ⁡ ( 7 5 ∘ ) = 3.106   kN . F v ​ = sin(75 ∘ ) F z ​ sin(30 ∘ ) ​ =3.106kN.