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)
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
sorry about repost
see below. I did that in my head by writing moment about center. One distance is twice as long.
For the first problem, we have a child sitting on a light swing and pulled back with a horizontal force. We need to find the tension force in each of the two supporting ropes.
To solve this, we can break down the forces acting on the swing. Let's assume that the angle between the swing ropes and the vertical direction is θ.
We know that the weight of the child is 160 N, acting vertically downwards. The pulling force is 100 N, acting horizontally backwards.
To find the tension force, we can resolve the forces into their components. The vertical component of the tension will counteract the weight, and the horizontal component will counteract the pulling force.
The vertical component of the tension will be equal to the weight of the child, which is 160 N. So we have:
Tension (vertical) = 160 N
Next, we need to find the horizontal component of the tension. We can use trigonometry to find this component. The horizontal component can be found using the equation:
Tension (horizontal) = Tension * cos(θ)
We know that the horizontal force is 100 N, so we can set up the equation:
100 N = Tension * cos(θ)
Now, we need to solve for the tension force. To do this, we can rearrange the equation:
Tension = 100 N / cos(θ)
To find the value of θ, we can use the relationship between the vertical and horizontal components of the tension:
Tension (vertical) / Tension (horizontal) = tan(θ)
Substituting the values we know:
160 N / (100 N / cos(θ)) = tan(θ)
Simplifying the equation:
160 N * cos(θ) / 100 N = tan(θ)
Now we have an equation with both cosine and tangent. To solve this, we can use a numerical method like iteration or use a graphing calculator to find the value of θ. Once we have the value of θ, we can substitute it back into the equation:
Tension = 100 N / cos(θ)
This will give us the tension force in each of the two supporting ropes.