Download Chris Khan 2007 Physics Chapter 6 FF represents the force of

Chris Khan 2007
Physics Chapter 6
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FF represents the force of friction, which slows down an object moving on any surface.
Kinetic Friction is the friction encountered when surfaces slide against one another.
µFN = FF, where µ represents the coefficient of kinetic friction.
o If I slide a 50g salt shaker across the table with an initial speed of 1.15 m/s and constant acceleration in
0.840 m, what is µ? First, draw a free body diagram. The forces acting on the salt shaker are: FN, FF,
and W (mg). First, let’s find a by doing v2 = vo2 + 2a(x – x0)  a = (v2 – vo2) / 2(x-xo) = -1.152 /
2(0.840) = -0.787 m/s2. Now, since ∑Fx = max = -FF (since that’s the only force in the negative xdirection), FF = -max. Next, ∑Fy = FN – W = may  FN = W = mg. Finally, since FF = µFN, µ = FF /
FN = -max / mg = -ax / g = -(-0.787) / 9.81 = 0.0802.
o If a sea lion slides down a 3m ramp 23 0 above the horizontal and µ = 0.26, how long does it take for
the sea lion to hit the water below the ramp? First, draw a free body diagram, which shows FN
perpendicular to the surface, W acting straight downward, F F at a slant up the ramp, and the x and y
vector components of each of these. ∑Fy = FN – Wy = may = FN – mg(cosΘ)  FN = mg(cosΘ). ∑Fx
= Wx - µFN = mg(sinΘ) - µ(mg(cosΘ)) = max. Solving for ax algebraically (in order to use it in x = ½
axt2), ax = g(sinΘ - µcosΘ) = (9.81)(sin23 – (0.26)cos23) = 1.5 m/s2. Now, since x = ½ axt2, t = √2x /
ax = √2(3) / 1.5 = 2s.
Static Friction keeps two surfaces from moving relative to one another.
µFN = FF where µ, here, represents the coefficient of static friction.
o If my truck’s ramp is 23.20 above the horizontal and my 95kg crate of baseball gloves starts to slide
down, what is µ in this case? First, draw the free body diagram, which shows F N perpendicular to the
surface, W (mg) going straight down, FF going up the ramp and the x and y vector components of each
of these. ∑Fy = FNy - Wy = FN – mg(cosΘ) = may = 0 (since there is no y-direction motion and FF does
not work in the y-direction). Therefore, FN = mg(cosΘ). ∑Fx = Wx - FFx = max = 0 = mg(sinΘ)- µFN
(since FF = µFN). *Note: Since we tilt the axis, we use sin for the x-direction and cos for the ydirection. Now, we can say ∑Fx = mg(sinΘ) - µFN = 0 and since we earlier said FN = mg(cosΘ), 0 =
mg(sinΘ) - µmg(cosΘ). If we solve algebraically for µ, µ = mg(sinΘ) / mg(cosΘ) = tanΘ = tan(23.2)
= 0.429.
FT represents the force of tension present in a string that is pulling an object.
o If a traction device like this is pulled by a mass (the box at the bottom) and the
force exerted on the circle by the middle pulley is 165N, what is the mass?
Firstly, ∑Fx = FTcos400 + FTcos400 = 2Tcos400. Also, ∑Fy = FTsin400 –
FTsin400 = 0, since the net force only acts in the x-direction when we tilt the
axis. Since 2Tcos400 = 165N, T = 165 / 2cos400 = 108N. Since F = ma, FT =
may = mg  m = T/g = 108 / 9.81 = 11kg.
Hooke’s Law tells us that Fx = -kx where a spring stretched or compressed by x from
its equilibrium length exerts a force given by Fx. Here, k is the force constant.
o If a 6.2kg pot hangs from two wires (one attached horizontally (string one)
and one sloping at a 400 angle (string two)), what is the FT in each wire?
Firstly, ∑FX = FT1x + FT2x = FT1 – T2cosΘ. Next, ∑Fy = FT1y + FT2y – W = 0 +
FT2sinΘ – mg = 0. From these two equations, FT1 = FT2cosΘ and FT2sinΘ =
mg  FT2 = mg / sin Θ = (6,2)(9.81) / sin400 = 94.6 N. Now, to find the
tension in string one, FT1 = FT2cosΘ = 94.6cos400 = 72.5 N.
When considering objects held together by a string, the FT in the string is the same for
both the x and y-direction since it’s the same string. If we find FT for box 1 and
then find FT for box 2, we can set the resulting equations equal to each other to
find the mass or acceleration or whatever we need to find.
o If m1 slides across a frictionless tabletop due to the pull of m2, what is a1,
what is a2, and what is FT? First, ∑Fx = FT = m1a and ∑Fy = W – FT =
m2g – FT = m2a. Therefore, FT = m1a and FT = m2g – m2a, so we can set
these equal: m1a = m2g – m2a.  m1a + m2a = m2g  a(m1 + m2) = m2g
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and therefore, a = m2g / m1 + m2. Lastly, FT = m1a = m1(m2g / m1 + m2), where what is in parenthesis
is what we found last step.
To make an object move in a circle with constant force, a force must act on it that is directed towards the center
of the circle. This means that the ball accelerates towards the center of the circle even though speed is constant
because acceleration is produced whenever the speed or direction of velocity changes. Here, direction changes
constantly. The center-seeking acceleration is known as centripetal acceleration.
When an object moves in a circle of radius r with constant speed v, its centripetal acceleration is: acp = v2/r.
A force must be applied to an object to give it circular motion. For an object of mass m, the net force acting on
it must have a magnitude given by Fcp = macp = mv2/r.
Here, fcp is centripetal force, which can be FT or FF, but varies in each system.
o If my Mercedes-Benz rounds the corner of radius 45 m and µ = 0.82, what is the greatest speed I can
go without skidding? ∑Fx = FF = max. Also, FF = µFN = macp (because of the circular path) = mv2/r.
Therefore, µFN = mv2/r. ∑Fy = FN – W = may = 0  FN = W = mg. Since FF = µFN and max = mv2/r,
µmg = mv2/r (since FN = mg, we can replace them). Now, to solve for v: v = √(µrg) = 19 m/s.
o Find the banking angle for a car traveling at 20.5 m/s in a turn of radius 85 m so that the car does not
need assistance from FF between the road and tires. ∑Fy = FNy – W = 0  FNy = W  FNcosΘ = W 
FN = W / cosΘ = mg / cosΘ. ∑Fx = FNx = FNsinΘ = max = mv2/r. Now, we can combine what we have
from both directions of motion to get: FNsinΘ = (mg / cosΘ)sinΘ = mv2/r (since FN = mg / cosΘ) 
tanΘ = v2/gr. Θ = tan-1(v2/gr)  tan-1((20.5)2 / (9.81)(85)) = 26.7o.
o While driving at 17 m/s, you encounter a dip with the radius of its circular arc at 65 m. What is F N
exerted by the seat on a 80 kg passenger at the bottom of the dip? First, ∑Fy = FN – mg = may = mv2/r
 FN = mv2/r + mg = m(v2/r + g) = 1140 N.