A chair of mass 8.8 kg is sitting on the horizontal floor; the floor is not frictionless. You push on the chair with a force of 56.0 N that is directed at an angle of 30.4 degrees below the horizontal and the chair slides along the floor. Calculate the normal force that the floor exerts on the chair.
56 . sin30.4? + 8.8 .9.8 = 114.58 = 110 N (Only 2 sf supported by data)
normal force + 56(sin(-30.4)) = 8.8kg (9.81) now solve that! remember everything has to be in equilibrium state in physics!!!!
When you push on the chair at 56 N, that push is both downward and horizontal...so by Newton's III Law (action = reaction, the floor is pushing back up an amount that not only equals the weight of the chair but the DOWNWARD component of the force you are applying...so a little reflection shows that the total downward force on the floor must be F(dnwd) = 56 sin 30.4 + 8.8 X 9.8 = 114.6 N so the floor has to be pushing straight upward the same amount. Remember that W = mg so you have to multiply the MASS 8.8 kg by the acceleration of gravity 9.8 m/s^2 to get the WEIGHT.
