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Part A

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A 10 kg box slides at a constant speed of 2 m/s along a smooth floor. What is the magnitude of the normal force exerted on the box by the floor?

Solution

We begin with a free body diagram for the box:

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From the free body diagram, we can write the equations of Newton's 2nd Law. We ignore the x-direction, since there are no forces acting.

\begin{large}\[ \sum F_{y} = N - mg = ma_{y}\]\end{large}

Because the box is sliding over level ground, it is not moving at all in the y direction. Thus, it certainly has no y-acceleration. Setting a_y = 0 in the above equation gives:

\begin{large}\[ N = mg = \mbox{98 N}\]\end{large}

Part B

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A person pushes a 10 kg box along a smooth floor by applying a perfectly horizontal force of 20 N. The box accelerates horizontally at 2 m/s². What is the magnitude of the normal force exerted on the box by the floor?

Solution

We begin with a free body diagram for the box:

Image Added

From the free body diagram, we can write the equations of Newton's 2nd Law.

\begin{large}\[ \sum F_{x} = F_{A} = ma_{x} \]
\[ \sum F_{y} = N - mg = ma_{y}\]\end{large}

Because the box is sliding over level ground, it is not moving at all in the y direction. Thus, it certainly has no y-acceleration. Setting a_y = 0 in the y direction equation gives:

\begin{large}\[ N = mg = \mbox{98 N}\]\end{large}

Part C

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A person is trying to lift a 10 kg box by applying a perfectly vertical force of 20 N with the help of a pulley. What is the magnitude of the normal force exerted on the box by the floor?

Solution

We begin with a free body diagram for the box:

Image Added

From the free body diagram, we can write the equations of Newton's 2nd Law. We ignore the x-direction, since there are no forces acting.

Part D

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A person pushes a 10 kg box along a smooth floor by applying force of 20 N. The force is applied at 30° below the horizontal. What is the magnitude of the normal force exerted on the box by the floor?

Solution

We begin with a free body diagram for the box:

Image Added

From the free body diagram, we can write the equations of Newton's 2nd Law.

\begin{large}\[\sum F_{x} = F_{A}\cos\theta = ma_{x}\]
\[ \sum F_{y} = N - mg - F_{A}\sin\theta = ma_{y}\]\end{large}

Because the box is sliding over level ground, it is not moving at all in the y direction. Thus, it certainly has no y-acceleration. Setting a_y = 0 in the y direction equation gives:

\begin{large}\[ N - mg -F_{A}\sin\theta = 0 \]\end{large}

Solving for the normal force gives:

\begin{large}\[ N = mg + F_{A}\sin\theta = \mbox{