A uniform beam of length 10m and mass 25kg is attached to a wall at a point A by a hinge at one end...?

Question

of the beam. A string is attached to the mid-point of the beam and to the wall at a point B on the wall (B is metres about A) so that the string is horizontal.

How do I calculate the tension in the string?

Appologies, allow me to correct the question, (I was writing too quick for my brain to keep up):

A uniform beam of length 10m and mass 25kg is attached to a wall at a point A by a hinge at one end of the beam. A string is attached to the mid-point of the beam and to the wall at a point B on the wall (B is 3 metres above A) so that the string is horizontal.

Calculate the tension in the string :)

Answer 1

I think you need to rephrase your question.
Well, based on my assumption here's the method:
Taking moments about A,
clockwise moments = anticlockwise moments
(5*250)Nm = (distance of B from A*tension in the string)Nm
I think the distance of B from A is provided in the question.

Answer 2

This problem is unsolvable as the distance between A and B is not a GIVEN. Nor is the gravitational constant. Is this happening on earth, the moon, in space? The equation in variables is:

Where:

L=length of beam (10 m)
m= mass of beam (25 kg)
x= distance from A to B (? m)
g= gravity (? m/s2)
N = Newtons force on string

N = ((L/2) / SQRT((L/2)^2-x^2)) * m * g

If this was occurring in space the answer would be 0 Newtons as g=0. On earth g= 9.81 m/s2

Now lets assume x=3 and g=9.81 m/s2

Then the answer is:

N = ((10/2) / SQRT((10/2)^2-3^2)) * 25 * 9.81 = 306.56 N

The trick is you have to draw the vector force triangle of the forces on the wall hinge. Then just use Pythagoras to solve for the vector mass on the string then multiply by gravity.

TK

nyphdinmd:

You were on the right track and another way to solve it. Good for you! What you missed was the string is horizontal and the beam is angled up.

Answer 3

You need to show the torques about the hinge sum to zero, then solve for teh tension.

The torque due to the weight of the beam is counterclockwise and is
Tb = l*m*g/2 where l = length of beam

The torque due to the string is

Ts = 1/2 l * T*sin(q) where T = tension and q = angle between string and beam.

Now you should be able to find q if you are given the height of the string above the hinge. Lets call that h, then

sin(q) = 2h/l

So Ts = 1/2*l*T*2h/l = T*h

Setting both toruqes equal (they have to sum to zero)

T*h = 1/2*m*g*l ---> T = m*g*l/(2h)

Answer 4

The twine's rigidity has a vertical and horizontal factor; yet for torque applications we purely care approximately its vertical factor (its horizontal factor factors right this moment on the hinge, and does not torque it the two way). So, if the vertical part of the rigidity is F, then you've 2 equations: F / sin 30° < 620 N F * 0.5 m = -? ? The sum of the torques in this final expression is non-trivial -- we don't definitely understand the place the middle of mass of the *sign* decrease than the beam is. yet enable's anticipate that it is likewise in the middle of the beam. Then the completed weight (-196 N - W) has an attempt arm of 0.5 m besides, and this 2nd equation reduces to purely: F = 196 N + W. wherein case we've: 196 N + W < 620 N * sin 30° W < 620 N sin 30° - 196 N.