Delphi51 said:The 66 degrees between the cable and the beam is not correct. Should be 66-39.
I would concentrate on the tip of the beam, drawing the 3 forces that act there. They must, of course, add up to zero. Unfortunately, I think two of the forces are unknowns so you are going to need a second equation. Torque?
I'm still getting 66-39 = 27 for the angle between the cable and the beam, where you have marked 37. The angle between the beam and horizontal is 66. From that you must subtract the 39 degrees between the cable and horizontal.a) the angle between the cable and the beam is 37 degrees and the angle
and the angle between the cable and the beam is the angle is where the pink arrow and star are in the picture, right? and that angle would be 14 degrees
I'm not following this. What is Fwall? If you mean the force with which the wall pulls on the rope, wouldn't that be T?b) Fx= Fwall - TCos(39) = 0
Fy= F(n of beam) - F(w beam) - TSin(39) - F(load) = 0
Okay, finding torques about the star point. Assuming the beam is free to turn about this point, so the total torque is zero. The weight of the beam is at an angle to the beam length L, and acts at its center of mass, so I think that torque is something likec) the reference point is where the purple star is, so
torque= -(L)F(weight of beam) + (L)F(normal of beam) - (L)F(w load) - (L)Tsin(39) +(L)F(n load) + T(load)
Delphi51 said:I'm still getting 66-39 = 27 for the angle between the cable and the beam, where you have marked 37. The angle between the beam and horizontal is 66. From that you must subtract the 39 degrees between the cable and horizontal.
I'm not following this. What is Fwall? If you mean the force with which the wall pulls on the rope, wouldn't that be T?
Okay, finding torques about the star point. Assuming the beam is free to turn about this point, so the total torque is zero. The weight of the beam is at an angle to the beam length L, and acts at its center of mass, so I think that torque is something like
0.5*L*cos(66)*(weight of beam).
I don't know what you mean by F . . .
We normally sum the forces on a point but these forces act far apart on the ends of the beam. I don't think we can say they total zero. The wall will push vertically as well as horizontally on the beam.Fx= Fwall - TCos(129) = 0
Fy= F(n of beam) - F(w beam) - TSin(39) - F(load) = ma
Delphi51 said:We normally sum the forces on a point but these forces act far apart on the ends of the beam. I don't think we can say they total zero. The wall will push vertically as well as horizontally on the beam.
It seems to me the sum of the forces on the tip of the beam would be zero, and the equation would be useful. Also, the torques about the beam's pivot point (if it is free to turn where it is attached to the wall).
A force is a push or pull that can cause an object to change its motion or shape. It is a vector quantity, meaning it has both magnitude and direction.
Forces can cause a cable or beam to bend, stretch, or compress, depending on the direction and magnitude of the force. These effects are known as tension, compression, and shear.
Torque is a measure of the twisting force that can cause an object to rotate. In the context of cables and beams, torque is important in determining the structural stability and integrity of the object.
The forces and torque on a cable or beam can be calculated using principles of mechanics, such as Newton's laws of motion and the equations of equilibrium. These calculations take into account factors such as the weight of the object, external forces, and the material properties of the cable or beam.
Understanding forces and torque on cables and beams is important in many fields, such as engineering, architecture, and construction. It is used to design and analyze structures, such as bridges, buildings, and mechanical systems, to ensure they can withstand the forces and torque they will experience in real-world situations.