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Part A

A person holds a 10 kg box against a smooth (i.e. frictionless) wall (as it slides down) by applying a perfectly horizontal force of 300 N. What is the magnitude of the normal force exerted on the box by the wall?

Solution

We begin with a free body diagram for the box:

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

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

Because the box is held against the wall, it has no movement (and no acceleration) in the x direction (a_x = 0). Setting a_x = 0 in the x direction equation gives:

\begin{large}\[ N = F_{A} = \mbox{300 N} \]\end{large}

Part B

A person moves a 10 kg box up a smooth wall by applying a force of 300 N. The force is applied at an angle of 60° above the horizontal. What is the magnitude of the normal force exerted on the box by the wall?

Solution

We begin with a free body diagram for the box:

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

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

Because Because the box is held against the wall, it has no movement (and no acceleration) in the x direction (a_x = 0). Setting a_x = 0 in the x direction equation gives:

\begin{large}\[ N = F_{A}\cos\theta = \mbox{150 N} \]\end{large}

Part C

A person scrapes a 10 kg box along a low, smooth ceiling by applying a force of 300 N at an angle of 30° above the horizontal. What is the magnitude of the normal force exerted on the box by the ceiling?

Solution

We begin with a free body diagram for the box:

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

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

Because Because the box is held against the ceiling, it has no movement (and no acceleration) in the y direction (a_y = 0). Setting a_y = 0 in the y direction equation gives:

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

which we solve to find:

\begin{large}\[ N = F_{A}\sin\theta - mg = \mbox{52 N}\]\end{large}