Download PWE 5-5: Pinned Against a Wall

Example 5-5 Pinned Against a Wall
You place a block of plastic that weighs 33.0 N against a vertical wall and push it
toward the wall with a force of 55.0 N (Figure 5-13). The coefficients of static and
kinetic friction between the plastic and the wall are ms = 0.420 and mk = 0.400,
respectively. Does the block remain at rest? If not, with what acceleration does it
slip down the wall?
Figure 5-13 ​​Pushing a block against a wall If you push a block against a wall with a force of a given
magnitude, will the block remain at rest or slide down the wall?
Set Up
Four forces act on the block: the gravitational
s block, the force Fsyou on block you exert
force w
to push it against the wall, the normal force
sblock that the wall exerts, and the upward
n
friction force sf (which opposes the downward
gravitational force). The block will only remain
at rest if the force of static friction is large
enough to balance the 33.0-N gravitational
force.
We’ll use Equation 5-1b to determine the
maximum force of static friction, which will
help us decide whether the block will slip. If
the block slips, we’ll use Equation 5-6 for the
force of kinetic friction to determine its
acceleration.
Solve
We use Newton’s second law to determine
the normal force that the wall exerts on the
block and Equation 5-1b to find the maximum
static friction force available. Only 23.1 N of
static friction is available, which is less than
the gravitational force on the block, so we
conclude that the block will slip.
Newton’s second law applied to the
block:
s
a Fext on block
s block + Fsyou on block + n
sblock + sf
= w
sblock
= mblocka
x
nblock
Fyou on block
wblock
Magnitude of the static friction force:
fs, max = msn
y
f
(5-1b)
Magnitude of the kinetic friction force:
fk = mkn
Newton’s second law in
component form for the block,
assuming the block remains at
rest (so the friction force is
static friction):
(5-6)
y
block at rest
fs,y = +fs
nblock,x = –nblock
x: Fyou on block + (2nblock) = 0
y: fs + (2wblock) = 0
wblock,y = –wblock
From the x component
equation,
nblock = Fyou on block = 55.0 N
From Equation 5-1b, the maximum static friction
force available is
fs, max = ms nblock = (0.420)(55.0 N) = 23.1 N
From the y component equation, the required static
friction force is
fs = wblock = 33.0 N
This force is more than the maximum static friction force
available, so the block can’t remain at rest.
x
Fyou on block,x
= +Fyou on block
We again apply Newton’s second law to the
block to find its acceleration as it slides down
the wall. Again, we find the normal force that
the wall exerts on the block, and we use this
value to find the kinetic friction force. Note
that we are given the block’s weight, not its
mass, so we have to calculate its mass mblock.
Newton’s second law in
component form for the block,
assuming the block is sliding
downward (so the friction force
is kinetic friction):
y
fk,y = +fk
nblock,x = –nblock
x: Fyou on block + (2nblock) = 0
y: fk + (2wblock) = mblock ablock, y
x
Fyou on block,x
= +Fyou on block
motion of block
wblock,y = –wblock
From the x component equation,
nblock = Fyou on block = 55.0 N
From Equation 5-6,
fk = mk nblock = (0.400)(55.0 N) = 22.0 N
Find the mass of the block:
wblock = mblock g
wblock
33.0 N
mblock =
= 3.37 kg
=
g
9.80 m>s 2
Substitute into the y equation and solve for ablock, y:
ablock, y =
Reflect
fk - wblock
22.0 N - 33.0 N
= -3.27 m>s 2
=
mblock
3.37 kg
We chose the positive y direction to be upward, so the negative value of ablock,y means that the block accelerates
downward. The magnitude of the acceleration is somewhat less than g = 9.80 m>s 2 because kinetic friction prevents
the block from attaining free fall.
Can you show that the block would remain at rest if you pushed with a force of at least 78.6 N?