A 160-N child sits on a light swing and is pulled back and held with a horizontal force of 100N. The tension force of each of the two supporting ropes is:
A uniform plank is supported by two equal 120N forces at X and Y at both ends of the plank.The support at X is then moved to Z (halfway to the plank center). The support of X and Z now have magnitudes of:
work for first problem:
weight is down, the pull is back, so the tension must act at an angle to counter both of these. The vertical component of the tension is 160N, the horizontal component is 100N.I understand the basic drawing but I am having trouble dealing with the angle.
work for the second problem:
ok so I said:
Fx + Fy= 240N
because both of them were 120N to be in equilibrium. I took the 240N to be the weight of the plank.
Then I moved Fx in my drawing to the Z part of the plank, halfway between the center of the board and the end. I then picked a length value to represent the distance. I choose 1m. So I said that Z must now be at 2.5m. I picked Fy as the torque point because it did not move and it was at the end.
so:
torque rotations about Fy
counterclockwise = clockwise
center of board weight down = Fz force up
(.5 X 240)= (.25 X Fz)
Fz= 480N (looks wrong)
Shouldnt the .25 x Fz be 2.5Fz? You said that was the distance, 2.5
I don't understand the problem diagram at all.
They didn't give a length so I thought I could just assume one. I assumed 10m. I wrote the wrong amount sorry. I got 48N, when I did the math. It doesnt match the choices given.
the choices are:
A) Fy=240N, Fz= 120N
B) Fy=40N, Fz=200N
c) Fy= 160N, Fz=80N
D Fy= 200N Fz=40N
e) Fy=80, Fz=160N
I drew something like this
original
Fx------------center-------------Fy
new:
------Fz------center-------------FY
Could you please help me just start off for the first problem.
Assume 240 at the center. Then Fz will be 2/3 of that, Fy will be one third.
Answer e
5 answers