202018 mass are suspended on epeoute sides of a ma om 19.7 kg hanging on apposit

Question

202018 mass are suspended on epeoute sides of a ma om 19.7 kg hanging on apposite sides o two objects the situation pictured in the Active upward, the other connecbed by an magnitude their accelerations must be of equal fThe abjects in the Awoed machine are subjact to the gavitational force a the strings connected to them. The free-body dagrams for the two abjects are shown in the Active nigure -0 forces act on each ceget: the upward force exerted by the streng and ne omard gravits such as one the p anis sesin th both sides of the pulley if the puley The signs ssed in problems such as this require care. When seject 1 acceleranes upware, ebject 2 accelerabes dowrward. Therer, ing up down should be represented equivalently by the same acceleration with the same sign. We can do hat by ining our sgn conventien wih up s positive for and down as postive for m wih this sign cenvention, the y component of the net force exerted on object 1 is T-and the y component of the net force exerted on object 2 is m Applying Newton's second law to mi gives the result T Do the same thing for The acceleration a is the same for both objects. When Equation (4) is added to Equation (2, T cancels and we obtaln Solving Equation (3) for«,and substituting for the given vales of -16.8 kg and m2-19.7 kg gives Substitute Equation (4) into Equation (1) to find Fiali" The acceleration given by Equation (4) can be interpreted as the otlo of the magnitude of the unbalanced fore (ma mJs to the total mass of the system, as expected from Newton's second lae. Notice thast the sign of the acceleration depends on the relative masses of the two oejects +m) 18284031 3/6

Explanation / Answer

4. form the given formula

ay = (m2 - m1)g/(m2 + m1)

now m1 = 16.8 kg

m2 = 19.7 kg

hence

g = 9.81 m/s/s

hence

ay = 0.779424 m/s/s

but

T = m1(g + ay)

hence

T = 177.9023342 N

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