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Question:

A chair of weight 125 N lies atop a horizontal floor; the floor is not frictionless.?

You push on the chair with a force of F = 45.0 N directed at an angle of 38.0^ below the horizontal and the chair slides along the floor.Using Newton's laws, calculate n, the magnitude of the normal force that the floor exerts on the chair. Express your answer in Newtons.

Answer:

The solution to this problem is the sum of the vertical forces exerted by the chair on the floor. This can be shown in the equation Fn = Fc + Fp where Fn = total normal force exerted by the chair on the floor, Fc = weight of the chair = 125 N, and Fp is the vertical component of force F applied to the chair. If you do a free body diagram and draw a force triangle you will see the following: cos(38°) = Fp / F; solve for Fp gives F * cos (38°) = Fp, putting in the numbers gives 45.0 N * cos(38°) = Fp 27.7 N = Fp Solve for Fn by using the known and calculated values as follows: Fn = Fc + Fp Fn = 125N + 27.7N Fn = 152.7N = the normal force that the chair exerts on the floor = the normal force that the floor exerts on the chair since the floor must exert a force that is equal to and opposite the normal force from the chair. I hope this helps.
Sep 30, 2017
Hmmm....this question means that g could be rounded to 10 m/s² ...yet as a results of fact it relatively is a factor of possibly a itemizing of such question possibly that information become pronounced at some past factor? The mass of the 50 N block = 50/g = 50/10 = 5 kg The tension pushing = 10 N The acceleration by Newton's 2nd regulation = tension/mass = 10/5 = 2 m/s² ANS
Sep 30, 2017
Chair Slides
Sep 30, 2017

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