A 20-kg child sits on a seesaw on the playground, 3.0 m from the
pivot. A 2nd child is seated on the other side of the pivot and 2.00 m
from the pivot. The mass of the second child needed to lift the first
child off the ground is?
A 20-kg child sits on a seesaw on the playground, 3.0 m from the pivot. A...
there are two children playing on a playground seesaw. one child has a mass of 25 kg the child sits on one end of the long uniform beam which has a mass of 20 kg. and another 45 kg child sits on the other end., the seesaw balances when a fulcrum is placed below the beam a distance of 1.50 m from the 25 kg-child, how long is the beam?
child a o child a Two children are balanced on a seesaw of mass 11.0 kg. The first child has a mass of 26.0 kg and sits 1.60 m away from the pivot. The center of mass of the seesaw is 0.113 m from the pivot. (a) How far from the pivot does the second child sit if he has a mass of 31.0 kg? (b) What is the supporting force exerted by the pivot? LPivot center of mass
A 32.0-kilogram child sits on a uniform seesaw of negligible mass; she is 2.10 m from the pivot point (or fulcrum). How far from the pivot point on the other side will her 22.0-kilogram playmate have to sit for the seesaw to be in equilibrium?
A child with a mass of 21 kg sits at a distance of 3 m from the pivot point of a seesaw. Where should a 15 kg child sit to balance the seesaw? m (from the pivot point)
At the local playground, a 21-kg child sits on the right end of a horizontal teeter-totter, 1.8 m from the pivot point. On the left side of the pivot an adult pushes straight down on the teeter-totter with a force of 151 N. Part A In which direction does the teeter-totter rotate if the adult applies the force at a distance of 3.0 m from the pivot? (clockwise/ counterclockwise) Part B In which direction does the teeter-totter rotate if the...
dont know what im doing wrong Suppose a 28.4-kg child sits 0.67 m to the left of center on the same seesaw as the problem you just solved in the PRACTICE IT section. A second child sits at the end on the opposite side, and the system is balanced. (a) Find the mass of the second child. 8.55 kg (b) Find the normal force acting at the pivot point. 362 Your answers to part (a) and (b) are not consistent....
Question 2 (2 Points) Static Equilibrium - Modified Seesaw (including mass of the seesaw) The two children shown in the Figure are balanced on a seesaw. The first child has a mass of m1= 16.2 kg and sits r1 = 2.1 m from the pivot. The center of mass of the seesaw is r3 = 0.36 m to the left of the pivot (on the side of the lighter child) and assume a mass of M = 13.4 kg for...
Suppose two children are using a uniform seesaw that is 3.00 m long and has its center of mass over the pivot. The first child has a mass of 30.0 kg and sits 1.40 m from the pivot. (a) Calculate where the second 18.0 kg child must sit to balance the seesaw. (b) What is unreasonable about the result? (c) Which premise is unreasonable, or which premises are inconsistent? Please show work and explanations.
II. Two children are sitting on a massless 4.0 m long seesaw which has a pivot at the middle point. If one child, with a mass of 36 kg sits 1.2 m from the pivot, where must the other child, of mass 54 kg, sit so that the seesaw remains in equilibrium? a. 0.6 m b. 0.8 m c. 1.2 m d. 1.4 m e. None of the above
Two children are balanced on a seesaw of negligible mass. The first child has a mass of 26 kg and sits 5m from the pivot. If the second child has a mass of 30kg, how far is she from the pivot?