Equilibrium of a Ladder: Finding Tension with a Weight of 100N

[jrd007]

Okay I cannot figure out the answers to any of these question.

1) A uniform steel beam has a mass of 940 kg. On it is resting half of an identical beam. What is the vertical support force at each end? Answers:5.8 x 10^3 N & 8.1 x 10^3N

------- (1/2)beam
--------------- 940 kg beam
|......|
|......|
|......| <~~~ supports


This problem does not give me a length at all, so I have no idea about how to go about solving this. All I am given is that the mass of the steel beam is 940 kg and the 1/2identical one is 470 kg. I have no idea how I can use my T1 + T2 + T3 = 0 or my F1 + F2 +F3 = 0 Equations. Can some one please help.

~~~~~~~~~~~~~

2) A 75 kg aduly sits at one end of a 9.0 m long board. His 25 kg child sits on the other end. (a)Where should the pivot be placed so the board is balanced? (b) Find the pivot point if the board is uniform and has a mass of 15 kg. Answers: (a)2.3 m (b)2.5 m

I am assuming you use the Torque equation? Someone please help me. :(

 

[Astronuc]

The sum of the forces of the supports must oppose the weights of the beams, so F_L + F_R = (940 kg + 470 kg) 9.8 m/sec².

Then pick either end and apply $\Sigma$M = 0.

Assume the moment arm of each beam is the distance between either support and the CM of the beam.

In problem 2, the same, the sum of the moments. If the board has length L, then one moment arm is x and the other is L-x.

 

[jrd007]

FL + FR = (940 kg + 470 kg) 9.8 m/sec2 = 13818 N

I picked one of the ends, so don't I just divided by 9.8 m/s^2 since I am looking for the weight? = 1,410 which is not correct. How do I set up the $\Sigma$M = 0.

 

[Astronuc]

OK, the big beam B1 has length L, and B2 (smaller beam) has length, L/2.

Pick the left side as the pivot. If the moment arms are located at the center of masses of each beam, what are the moment arms, x1 and x2, of B1 and B2, respectively.

The Moments are just the products of the forces and moment arms.

So -W1*x1 - W2*x2 + ?? = 0, where W1 is weight of B1 and W2 is weight of B2.

I use the convention that a positive moment would cause a counterclockwise rotation.

 

[jrd007]

W1*x1 - W2*x2 + ?? = 0 that gives us 4606...

 

[Astronuc]

What did you use for x1 and x2?

 

[jrd007]

I do not know. What do I use for x1 and x2?

 

[jrd007]

No length is given so I do not know how we are expected to get a position, x.

 

[Astronuc]

One does not need a numerical value for this problem.

Take the length of the long beam (940 kg) as L, and so the half beam has length L/2.

Now, taking the left side as 0, what the is the location of the CM's of the long beam and the half beam? This will give the moment arms about the left support for the long beam and half beam. The moment arm of the right support is L.

 

[marlon]

Okay I cannot figure out the answers to any of these question.
1) A uniform steel beam has a mass of 940 kg. On it is resting half of an identical beam. What is the vertical support force at each end? Answers:5.8 x 10^3 N & 8.1 x 10^3N
------- (1/2)beam
--------------- 940 kg beam
|......|
|......|
|......| <~~~ supports
This problem does not give me a length at all, so I have no idea about how to go about solving this. All I am given is that the mass of the steel beam is 940 kg and the 1/2identical one is 470 kg. I have no idea how I can use my T1 + T2 + T3 = 0 or my F1 + F2 +F3 = 0 Equations. Can some one please help.
~~~~~~~~~~~~~
2) A 75 kg aduly sits at one end of a 9.0 m long board. His 25 kg child sits on the other end. (a)Where should the pivot be placed so the board is balanced? (b) Find the pivot point if the board is uniform and has a mass of 15 kg. Answers: (a)2.3 m (b)2.5 m
I am assuming you use the Torque equation? Someone please help me. :(

For the first question, keep it simple. Just assume that the length of the longest bar is l (you do not need to know this value).

You have 4 forces : F+F' - 4700 - 9400 = 0
I took 10 m/s² for g and F is the upward (along Y axis) support force on the left side, F' is the one on the right hand side.

Now, we calculate the moments (+ for counterclockwise rotation). The clue is to take the pivot point in the origin (there were the left hand side support force F originates). F originates in the origin, 4700 N at distance l/4, 9400N at l/2 and F' at l.

Now, what is the equation for the sum of the moments in this case ?

Once you have that, you will be able to get rid of the distance l and together with the equation for the forces, you have in total 2 equations with 2 unkowns F and F'

marlon

 

[jrd007]

That is the problem I am having. Forming the sum of the moments equations... would it be 4700 N + 9400 N + F = 0

 

[marlon]

That is the problem I am having. Forming the sum of the moments equations... would it be 4700 N + 9400 N + F = 0

Not at all. Do you know what a moment is ?

It is defined as the vector product of vector r and force-vector F. Vector r denotes the position where the force originates. Now, in your case, you are lucky because the position vector r and the force F are perpendicular to each other. So, you can just multiply the two. beware of the signs, though, make sure you follow the convention (+ for counterclockwise rotation). For each force, i already gave you the r vector...

marlon

 

[jrd007]

Okay, can you atleast set up the equation for me? Maybe that will help me better understand? I am lost.

 

[Doc Al]

1) A uniform steel beam has a mass of 940 kg. On it is resting half of an identical beam. What is the vertical support force at each end? Answers:5.8 x 10^3 N & 8.1 x 10^3N

------- (1/2)beam
--------------- 940 kg beam
|......|
|......|
|......| <~~~ supports

This problem does not give me a length at all, so I have no idea about how to go about solving this. All I am given is that the mass of the steel beam is 940 kg and the 1/2identical one is 470 kg. I have no idea how I can use my T1 + T2 + T3 = 0 or my F1 + F2 +F3 = 0 Equations. Can some one please help.

To analyze this problem (as others have explained) start by identifying the forces acting on the beams. I see 4 forces:
(1) the weight of the 940 kg beam (where does it act?)
(2) the weight of the 470 kg beam (where does it act?)
(3) the upward force on the left support; call it F1
(4) the upward force on the right support; call it F2

Now apply the conditions for equilibrium. First the force equation: The net force must add to zero. Set up that equation using the four forces above.

Then the torque equation: The net torque about any point must add to zero. Torque is defined as Force X perpendicular distance to the axis. Choose the left support as your axis. What's the torque due to force (1) above about that axis? (Don't get hung up on the length of the beam: Call it L. It will cancel out.) Call torques that tend to rotate the beams in one direction to be positive and the other direction to be negative. (It doesn't matter which is which.) Find the torques due to all 4 forces (about the left axis) and set up the torque equation. Try it.

 

[marlon]

Okay, can you atleast set up the equation for me? Maybe that will help me better understand? I am lost.

i gave you nearly everything man. I gave you the formula to calculate each moment separately. Just apply this formula for each force and add up the 4 momenta you will get (one of them will be 0).

Besides, read Doc Al's last post in this thread and think things over. It is really very simple.

 

[jrd007]

I get it now, Doc! Thanks.

It would be (1/2)mg(1/2)L + mg(1/2L) = F2(L)
L cancel out. So (5/8)mg = F2
F2 = 5757

Then you use F1 = F(m) + F(1/2m) - F2
And you get F1 = 8096! Thanks for all the help everyone.

Now solving for where to place my pivot point, would I do something like
F1(adult) + F2(child) = 0 (but where do I put my x = pivot point)?

 

[Doc Al]

Now solving for where to place my pivot point, would I do something like
F1(adult) + F2(child) = 0 (but where do I put my x = pivot point)?

You need to solve for the point where the torques (from the three forces acting) would balance. (Call the pivot point a distance X from one end and solve for it by writing the torque equation about that pivot point.)

 

[jrd007]

T1 + T2 + T3?

but in the first one we do not know the weight of the board. So would it be

(75 kg x 9.8 x 1m) + (25 kg x 9.8 x 9m) + F(x) ?

 

[Doc Al]

Even though it is not clear from the way the problem is worded, I believe that for part (a) they want you to ignore the mass of the board. Thus for part (a) only two forces act: the weights of the adult and child. To find the balance point, set up the torque equation so that clockwise torques balance counterclockwise torques. Call the distance from the adult to the pivot point x. (What's the distance from the child to the pivot point?) Answer these questions:

(1) What's the torque due to the adult's weight?
(2) What's the torque due to the child's weight?

Set them equal and solve for x.

 

[jrd007]

The adults T would be (75 kg x 9.8 m/s^2)(x) = (25 kg x 9.8 m/s^2)(childs distance)
Would the child be 9 m away or 4.5?

 

[Doc Al]

Would the child be 9 m away or 4.5?

Neither. Draw yourself a picture. On the left (say) is the adult; on the right, 9 m away from him, is the child. The pivot point is between them, a distance x from the adult. So what's the distance between pivot point and child?

 

[Astronuc]

If one has board of length a, and one tries to find a pivot point x, then the other side must be a-x.

For the first part, ignore the mass of the board. In the second part, the boards weight acts at the center of mass on that side of the board, as in the beam problem.

If one has lengths x and a-x, then the respective CMs are x/2 and (a-x)/2, assuming the cross-section or linear mass is constant.

 

[jrd007]

(75 kg x 9.8 m/s^2)(x) = (25 kg x 9.8 m/s^2)(9-x) ?
735(x) = 245(9-x)
735x = 2205 - 245x
980x = 2205
x = 2.25 = 2.3 which is the correct answer.

And if the board is acting w/ a 15 kg mass then...

(75 kg x 9.8 m/s^2)(x) = (25 kg x 9.8 m/s^2)(9-x) ?

 

[Doc Al]

And if the board is acting w/ a 15 kg mass then...

Then there are 3 forces (and torques) to consider. You have to add the torque due to the weight of the board.

 

[jrd007]

(75 kg x 9.8 m/s^2)(x) + (25 kg x 9.8 m/s^2)(9-x) + (15kg)(9.8 m/s)(1/2 x) ?

 

[Doc Al]

Two problems:
(1) Since the midpoint of the board is 4.5 m from the end, the distance between midpoint and pivot is "4.5 - x".
(2) Realize that some of these torques act clockwise, while some act counterclockwise. Give them different signs accordingly.

 

[jrd007]

Th adult's T will be ccw, where as the other 2 are cw, right?

(-75 kg x 9.8 m/s^2)(x) + (25 kg x 9.8 m/s^2)(9-x) + (15kg)(9.8 m/s)(4.5 - x)

 

[Doc Al]

Th adult's T will be ccw, where as the other 2 are cw, right?

That's right.

 

[jrd007]

thanks. It worked!

 

[kingfiz]

can somebody help..
the question:weight of the ladder is 100N.assume the system is frictionless.find the tension.