# What is the magnitude of the acceleration of the chair, and the magnitude of the normal force acting on the chair in the followig case: Yusef pushes a chair of mass 45.0 kg across a carpeted floor with a force of magnitude 144 N directed 35.0 degrees below the horizontal. The magnitude of the frictional force between the carpet and the chair is 89.0 N.

Yusef pushes the chair of mass 45 kg across a carpeted floor with a force of magnitude 144 N directed 35.0 degrees below the horizontal. The magnitude of the frictional force between the carpet and the chair is 89.0 N.

The horizontal component of the force applied on the chair by Yusef that causes it to move forward is equal to 144*cos 35 = 117.95 N. There is a force of 89 N due to the friction between the chair and the carpet acting in the opposite direction. The net horizontal force is 117.95 - 89 = 28.95 N. This force accelerates the chair with a mass of 45 kg by 0.6435 m/s^2.

The normal force acting on the chair is the sum of the weight of the chair and the vertical component of the force applied by Yusef on the chair. This is equal to 45*9.8 + 144*sin 35 = 523.59 N

The acceleration of the chair is 0.6435 m/s^2 and the normal force is 523.59 N.

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