ok so i need a lil help with this, if..A chair of mass 10.5 kg is sitting on the horizontal floor; the floor is not frictionless. You push on the chair with a force = 40.0 N that is directed at an angle of 39.0 degrees below the horizontal and the chair slides along the floor.Calculate the normal force that the floor exerts on the chair.ive tried a couple times and can tell you the answer is NOT, 103 N, 367 N, 429 N and 77.8 N.Please help me out and include the CORRECT answer and working out.thanks heaps, take it easy :)
Weight of chair = mg = 10.5 x 9.81 = 103.0N Vertical component of 40.0N force = 40.0sin(39.0) = 25.17N upwards Resultant vertical force on chair = 0 (as there is no acceleration in vertical direction), therefore: total downwards force = total upwards force The total downwards force is just the weight. The total upwards force is 25.17N + the normal force. 103.0 = 25.17 + N where N is normal force N = 103.0 - 25.17 = 77.8N (to 3 sig. figs., same as data supplied)
