- #1
blackout85
- 28
- 1
Torque
A 160 N child sit on a light swing and is pulled back and held with a horizontal force 100 N. 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:
a) the first one I don't know where to begin. I know the child's weight acts down 160N and that the tension of the ropes would have to support the child's weight. There would also be a 100N force acting up. I can't do this equilibrium way so I am lost.
b) 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)
then I am stuck...
Homework Statement
A 160 N child sit on a light swing and is pulled back and held with a horizontal force 100 N. 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:
Homework Equations
a) the first one I don't know where to begin. I know the child's weight acts down 160N and that the tension of the ropes would have to support the child's weight. There would also be a 100N force acting up. I can't do this equilibrium way so I am lost.
b) 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)
then I am stuck...