Why does a long pole help a tightrope walker stay balanced?

Please show me how to calculate where I need to place the stack of bricks on the right side in order for the beam to become balanced? Please refer to image.

The diagram above shows a person walking along a uniform beam. The beam has a mass of 295 kg and is supported in the two locations shown. The person has a mass of 115 kg. How close can the person get to the right-hand end of the beam without causing it to rotate?

It is the property of the body to resist a rotation. A Moment of Inertia (B Rotational KE C Angular Momentum D Moment of torque

A cat of mass M and a bowl of tuna fish of mass 2M/5 are at opposite ends of a seesaw of length L = 4 meters. The cat is a distance d from the pivot as indicated on the figure. How far to the left of the pivot must a cat of mass 4M/5 stand to keep the seesaw balance?

(The complete question is in the picture) A meter stick balances horizontally on a knife-edge at the 50.0 [cm]mark. If you put a stack of 15.0 [g] coins over the 0.100 [m] mark and a stack of 5.0 [g] coins over the 0.750 [m] mark, the stick is found to balance at the 0.45 [m] mark. What is the force exerted by the knife-edge to the meter stick?A. 0.735 [N]B. 0.931 [N]C. 1.52 [N]D. 2.55 [N]

When an acrobat walks on a tightrope, she extends her arms straight out from her sides. She does this to make it easier for her to catch herself if she should tip to one side or the other. Explain how this works.

A rigid body is rotating. Is it necessary that a net torque acts on the body?

Consider a teeter-totter that is made out of pine, shown below. A pineapple sits on one side of the teeter-totter, a distance 1.09 m from the pivot point, while an apple (non-pine) sits on the other side, a distance of 2.63 m from the pivot. If the pineapple has a mass 132 g, what must the mass of the apple be in order to balance the teeter-totter (i.e. to maintain rotational equilibrium)? [Answer in units of grams (g) with 3 sig figs but do not enter units with your answer]

A hanging weight, with a mass of m, 0.355 kg, is attached by a rope to a block with mass m₂ - 0.815 kg as shown in the figure below. The rope goes over a pulley with a mass of M-0.350 kg. The pulley can be modeled as a hollow cylinder with an inner radius of R, 0.0200 m, and an outer radius of R₂ = 0.0300 m; the mass of the spokes is negligible. As the weight falls, the block slides on the table, and the coefficient of kinetic friction between the block and the table is = 0.250. At the instant shown, the block is moving with a velocity of v, - 0.820 m/s toward the pulley. Assume that the pulley is free to spin without friction, that the rope does not stretch and does not slip on the pulley, and that the mass of the rope is negligible. R₂ R (a) Using energy methods, find the speed of the block (in m/s) after it has moved a distance of 0.700 m away from the initial position shown. m/s (b) What is the angular speed of the pulley (in rad/s) after the block has moved this distance? rad/s