Balancing Two Uniform Beams: Calculating Forces & Angles

[masamune]

Two identical uniform beams, each having a mass of 4 kg, are symmetrically set up against each other on a floor with which they have a coefficient of static friction µ = 0.609 .

a) What is the minimum angle the beams can make with the floor without falling?
b) For the angle calculated in part (a), what is the magnitude of the force exerted on one beam by the other beam, at the apex?

Next a heavy point mass 1.2 kg is balanced on the apex.
c) What is the normal force exerted on each beam by the floor?
d) For this case, what is the minimum angle the beams can make with the floor without sliding?

Thanks for the help in advance...see the attached picture for what the problem looks like.

 

[cuz937100]

I'm having problems with this same problem too. This is what the help says for the first two parts...

HELP: Draw a FBD and make sure all forces and torques sum to zero.
HELP: Your FBD should show 4 different forces on each beam. A convenient point about which to calculate the torque for one beam is the point where the beam rests on the ground.

HELP: Consult your FBD and examine the horizontal forces.

I tried and I think the four forces are acting on each beam are the (weigh, normal, force of friction, and the force from the opposing beam). Can someone help out?

 

[M98Ranger]

a) To find the minimum angle the beams can make without falling, we need to consider the forces acting on the beams. The weight of each beam will act downwards, while the normal force from the floor will act upwards. The beams will also exert a force on each other at the apex. The equilibrium condition for the beams to not fall is when the net force and net torque on each beam is zero.

Using the equation for torque, τ = rFsinθ, we can calculate the torque exerted by the weight of the beam and the force at the apex. Since the beams are symmetrically set up, the torque from the weight will be equal and opposite on each beam, cancelling out. The torque from the force at the apex will also be equal and opposite on each beam, cancelling out. Therefore, the minimum angle the beams can make without falling is when the net torque on each beam is zero, which occurs when the beams are perfectly horizontal.

b) At the apex, the beams will exert a force on each other that is equal and opposite. This force can be calculated using the equation F = µN, where µ is the coefficient of static friction and N is the normal force. Since the beams are symmetrically set up, the normal force on each beam will be equal, and therefore the force exerted on one beam by the other will also be equal. Plugging in the given values, we get F = (0.609)(4 kg)(9.8 m/s^2) = 23.8 N.

c) With the addition of the heavy point mass, the normal force exerted by the floor on each beam will increase. This is because the weight of the point mass will also be supported by the beams, and therefore the normal force from the floor will be the sum of the weight of the beams and the point mass. Using the equation F = mg, where m is the mass and g is the acceleration due to gravity, we can calculate the normal force on each beam to be N = (4 kg + 4 kg + 1.2 kg)(9.8 m/s^2) = 88.4 N.

d) To find the minimum angle the beams can make without sliding, we need to consider the forces acting on the beams again. In addition to the weight and normal force, there will also be a frictional force acting on each beam. This frictional force can be calculated using the