# 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

## To solve the first problem about the tension force in the swing ropes, we can break down the given forces into their vertical and horizontal components.

Given:

Weight of child = 160 N (downward)

Pull-back force = 100 N (horizontal)

Step 1: Draw a diagram representing the forces acting on the swing. Label the forces and their directions.

Step 2: Decompose the forces into their vertical and horizontal components. Assuming the angle between the two ropes is θ:

Vertical component of tension (Tv) = Weight of child = 160 N

Horizontal component of tension (Th) = Pull-back force = 100 N

Step 3: Use trigonometric functions to find the magnitudes of the tension forces.

Since we have the vertical and horizontal components of tension, we can use the following trigonometric relationships:

sin(θ) = Tv/T

cos(θ) = Th/T

Where T is the magnitude of the tension force.

Step 4: Solve for the magnitude of the tension force (T) using the given components.

From the vertical component:

sin(θ) = Tv/T

sin(θ) = 160/T

T = 160/sin(θ)

From the horizontal component:

cos(θ) = Th/T

cos(θ) = 100/T

T = 100/cos(θ)

Step 5: Set the two expressions for T equal to each other and solve for θ.

160/sin(θ) = 100/cos(θ)

Step 6: Once you solve for θ, substitute the value back into either of the equations from Step 4 to find the magnitude of the tension force (T).

This method will give you the angle and magnitude of the tension force, which can then be used to answer the question.

For the second problem about the support forces on a plank, it seems that you are on the right track. However, there appears to be some confusion in your calculations.

Given:

Force at X = Force at Y = 120 N

Support at X is moved to Z (halfway to the plank center)

Assuming the weight of the plank is 240 N, we need to determine the forces at X and Z.

Step 1: Draw a diagram representing the forces acting on the plank. Label the forces and their directions.

Step 2: Apply the conditions for equilibrium, which state that the sum of all forces acting on the plank should be zero.

Sum of vertical forces = 0

Sum of torques (rotational forces) = 0

Step 3: Calculate the sum of vertical forces to find the support force at Z.

Since both X and Z contribute equal upward forces, and the weight of the plank is downward, we have:

Support force at Z (Fz) + Support force at X (Fx) - Weight of plank = 0

Step 4: Calculate the sum of torques to find the support force at Y.

Taking the support at Y as the pivot point, the torque due to the weight (W) must be balanced by the torque from the support force at Z.

Torque due to weight (W) = Torque due to support force at Z

Step 5: Solve the equations obtained in Steps 3 and 4 to find the magnitudes of Fz and Fy.

With the given information, substitute the values into the equations and solve for Fz and Fy.

Step 6: Compare the calculated values with the given options to find the correct combination of forces at Z and Y.

It seems that there was a mistake in your calculations, which resulted in an incorrect answer. By following the steps outlined above, you should be able to find the correct magnitudes of the support forces at Z and Y.