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
System:
Interactions:
Model:
Approach:
Diagrammatic Representation
We begin with a free body diagram for the box:
It is important to note that any surface has the potential to exert a normal force and that the normal is always perpendicular to the plane of the surface. If the wall did not exert a normal force, the box would simply pass through it.
Mathematical Representation
From the free body diagram, we can write the equations of Newton's 2nd Law.
\begin
[\sum F_
= F_
- N = ma_
]
[ \sum F_
]\end
Because the box is held against the wall, it has no movement (and no acceleration) in the x direction (ax = 0). Setting ax = 0 in the x direction equation gives:
\begin
[ N = F_
= \mbox
]\end
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
System:
Interactions:
Model:
Approach:
Diagrammatic Representation
We begin with a free body diagram for the box:
Mathematical Representation
From the free body diagram, we can write the equations of Newton's 2nd Law.
\begin
[\sum F_
= F_
\cos\theta - N = ma_
]
[ \sum F_
= F_
\sin\theta - mg = ma_
]\end
Because Because the box is held against the wall, it has no movement (and no acceleration) in the x direction (ax = 0). Setting ax = 0 in the x direction equation gives:
\begin
[ N = F_
\cos\theta = \mbox
]\end
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
System:
Interactions:
Model:
Approach:
Diagrammatic Representation
We begin with a free body diagram for the box:
The ceiling must push down to prevent objects from moving up through it.
Mathematical Representation
From the free body diagram, we can write the equations of Newton's 2nd Law.
\begin
[\sum F_
= F_
\cos\theta = ma_
]
[ \sum F_
= F_
\sin\theta - mg - N = ma_
]\end
Because Because the box is held against the ceiling, it has no movement (and no acceleration) in the y direction (ay = 0). Setting ay = 0 in the y direction equation gives:
\begin
[ F_
\sin\theta - mg - N = 0 ]\end
which we solve to find:
\begin
[ N = F_
\sin\theta - mg = \mbox
]\end
We can check that the y direction is in balance. We have N (52 N) and mg (98 N) on one side, and FA,y on the other (150 N).