Practice 4.1 Upper Tenmon Cable crane is used to hoist a load of mass m, = 500 kilograms. The load is suspended by a cable from a hook of mass m2 # 50 kilograms, as shown in the diagram above. The load is lifted upward at a constant upward acceleration of 2 m/s?. 2 2 = 50 kg (a) On the diagrams below draw and label the forces acting on the load and the forces acting on the hook as they accelerate upward. The size of your force vectors should reasonably reflect the relative Lower Tennon magnitude of the forces. Cable Load Hook Load m, =500 kg (b) (b) has two parts. Each assesses the same physics, but (i) is variable-only (like we practice in class) and (ii) lets you plug in numbers. If you feel stuck/confused on part (i), skip it and solve part (ii) as you normally would and you will still earn most of part (b)'s credit. (1) Beginning with Newton's Laws, derive an equation for the tension in the lower cable in terms of m, my a, and g. Show your process of developing the equation. Hint: "in terms of" means that your final answer can only use those variables. You do have numbers for m, my, a, and g, but part (i) is about making a generic equation that would apply to any similar situation. (ii) Use your expression from part (b)(i) to calculate the tension 7, in the lower cable. You don't need to redo any work from (b)(i)! If you have a finished variable expression from (i) just plug in and solve! (c) Calculate the tension 72 in the upper cable