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A 475 kg piano is being lowered into position by a crane whiletwo people steady

ID: 1666463 • Letter: A

Question

A 475 kg piano is being lowered into position by a crane whiletwo people steady it with ropes pulling to the sides. Bob's ropepulls to the left, 15° below horizontal,with 535 N oftension. Ellen's rope pulls toward the right, 25° belowhorizontal.
(a) What tension must Ellen maintain in her rope to keep thepiano descending at a steady speed?
(b) What isthe tension in the main cable supporting the piano?
So Iunderstand that And to maintain a steadyspeed acceleration along the y axis must be 0. What I am troubled with isif the downward force is 4659.75 N (475*9.81) then the upward forceexerted by the rope has to be 4659.75 N then this must be split between the two pullers. Can someone tell me if my infois wrong so, could start me in the right direction, thankyou.
(a) What tension must Ellen maintain in her rope to keep thepiano descending at a steady speed?
(b) What isthe tension in the main cable supporting the piano?
So Iunderstand that And to maintain a steadyspeed acceleration along the y axis must be 0. What I am troubled with isif the downward force is 4659.75 N (475*9.81) then the upward forceexerted by the rope has to be 4659.75 N then this must be split between the two pullers. Can someone tell me if my infois wrong so, could start me in the right direction, thankyou. And to maintain a steadyspeed acceleration along the y axis must be 0. What I am troubled with isif the downward force is 4659.75 N (475*9.81) then the upward forceexerted by the rope has to be 4659.75 N then this must be split between the two pullers. Can someone tell me if my infois wrong so, could start me in the right direction, thankyou.

Explanation / Answer

The horizontal components are supplied by the people and forthe piano not to move to either side, these components must beequal. By doing this, you can find the tension in Ellen'srope. cos = adj/hyp: adj = cos *hyp = cos 15 * 535= 516.77N to the left. So the x-component must be 516.77 to the right for the otherperson. If they are pulling at 25 degrees, the equationis cos = adj/hyp: hyp = adj/cos = 516.77N/cos25= 570.2N - this is the tension for ellen Now, the two people are pulling downward with the piano andyou need to account for this force downward along with the weightof the piano sin 15 * 535N = 138.47N sin 25 * 570.2N = 240.98N Now the total downward force is equal to 138.47 + 240.98 +4659.75 = 5039.2N If it is moving at constant speed, the downward weight mustequal the upward tension in the main cable. Hope this helps
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