Question

A chair of mass 11.5 kg is sitting on the horizontal floor; the floor is not...

A chair of mass 11.5 kg is sitting on the horizontal floor; the floor is not frictionless. You push on the chair with a force F = 42.0 N that is directed at an angle of 38.0 below the horizontal and the chair slides along the floor.

*Use Newton's laws to calculate the normal force that the floor exerts on the chair.
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Answer #1
Concepts and reason

The concept used to solve the problem is Newton’s second law and equilibrium condition for force.

Sketch the free body diagram of the problem and specify the normal force, weight force, and the friction force.

Finally, calculate the normal force from the diagram using the newtons law.

Fundamentals

Newton’s Second Law: The Newton’s second law states that the net force on an object is the product of mass of the object and final acceleration of the object. The equation of the Newton’s second law is,

F=ma\sum {\vec F = m\vec a}

Here, F\sum {\vec F} is the net force on the object, mm is mass of the object, and a\vec a is the acceleration of the object.

The equilibrium condition for the force gives,

ΣF=0\Sigma F = 0

Here, FF is force.

The weight force is expressed as follows:

W=mgW = mg

Here, m is the mass, g is the gravity, and W is the weight force.

Consider the free body diagram of the problem shown below:

-
F cose<
28.00
-
sin o
mg

The above picture shows the normal force NN , applied force FF , and weight force mgmg .

The normal force is calculated using the newtons law as follows:

N=W+FsinθN = W + F\sin \theta

Here, N is the normal force, W is the weight force, F is the force, and θ\theta is the angle of the slope.

Substitute mg for W in the above expression.

N=mg+FsinθN = mg + F\sin \theta

Substitute 11.5 kg for m, 9.8m/s29.8\,{\rm{m/}}{{\rm{s}}^{\rm{2}}} for g, 42.0 N for F, and 38.038.0^\circ for θ\theta in expression N=mg+FsinθN = mg + F\sin \theta .

N=(11.5kg)(9.8m/s2)+(42.0N)sin38.0=138.56N\begin{array}{c}\\N = \left( {11.5\,{\rm{kg}}} \right)\left( {9.8\,{\rm{m/}}{{\rm{s}}^{\rm{2}}}} \right) + \left( {42.0{\rm{N}}} \right)\sin 38.0^\circ \\\\ = 138.56\,{\rm{N}}\\\end{array}

Ans:

The magnitude of normal force that the floor exerts on the chair is 138.6 N.

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