1G10-002Dogsledteam–Newton’sSecondLaw–PROMPTSAFTER
STATEMENTOFTHEPROBLEM
y
a
Ateamofdogsispullingtwoconnectedsledswitha
x
constantaccelerationof2.3m/s2.Thepassengersled,
connectedtothedogsinfront,hasamassof96kg.Thecargosled,tiedtothefrontsled,has
Passenger Sled
Cargo Sled
amassof42kg.Assumefrictionbetweenthesledsandsnowisnegligible.
n
n
Fpc
Fd 𝐹! ?
Fcp
a)Howmuchistheforcethatthedogteamexertsonthesledtrain,
Fcf
b)Whatisthemagnitudeoftheforcethatthecargosledexertsonthepassengersled,𝐹!" ?
Fpf
Fg
Fg
STRATEGY
a = 2.3 m/s2
m
= 96 kg
The accelerating sleds are subject to Newton’s Second Law (N2L), which states that the acceleration of passenger
an
=
42
kg
µ
=
0.007
object is proportional to the net applied force and inversely proportional m
to the
mass
of
the
accelerating
cargo
object. We need to draw free-body diagrams of each object of interest, sum the forces acting on each
Fd = ?
Fpc = ?
object, and solve for the unknown force using N2L, 𝐹 = 𝑚𝑎 . We will also use Newton’s Third Law
which states that when two objects are connected and the first one exerts a force on the second one, the
second one responds with a reaction force of the same magnitude, acting back on the first one.
First, we draw free-body diagrams for each object of interest in the problem. There are three objects of
interest: 1) the sled train, 2) the passenger sled, and 3) the cargo sled. Representing each
object
as a point,
Sled
Train
y The forces are defined as
we draw all the forces acting on each object and define a coordinate system.
n
follows:
Both Sleds
Fd
Free-BodyDiagrams(FBDs)
𝑭𝒅 is the force exerted by the dogs on the sled train.
x
𝑭𝒄𝒑 is the force exerted by the cargo sled on the passenger
Fg
sled.
Passenger Sled
Cargo Sled
𝑭𝒑𝒄 is the force exerted by the passenger sled on the cargo
sled.
n
n
Fcp
Fd
Fpc
𝑭𝒈 is the force exerted by gravity on the sled train, also
called weight (𝐹! = 𝑚𝑔 )
𝒏 is the force exerted by the ground on each sled, called
the normal force. Now we can sum the forces acting on the sled train,
𝐹 = 𝑚𝑎, and solve for 𝐹! . Since we know the mass
and the acceleration of the cargo sled, we can also
determine the force exerted by the passenger sled on the
cargo sled, 𝐹!" . Applying Newton’s Third Law, the force
exerted by the cargo sled on the passenger sled (𝐹!" ) has
the same magnitude as the force exerted by the passenger
sled on the cargo sled (𝐹!" ).
CALCULATION
The net force on the sled train in the x-direction is:
𝐹! = 𝑚𝑎! Fg, c
Fg, p
a = 2.3 m/s2
mcargo = 42 kg
Fcp = ?
mpassenger = 96 kg
Fd = ?
1G10-002Dogsledteam–Newton’sSecondLaw–PROMPTSAFTER
𝐹! = 42 kg + 96 kg (2.3 𝑚/𝑠 ! )
𝐹! = 𝑚𝑎! The dog team pulls the sled train with a force 𝐹! =
320 𝑁 in the +x-direction
The net force on the cargo sled in the x-direction is:
𝐹! = 𝑚𝑎! 𝐹!" = 𝑚𝑎! !
𝐹!" = 42 kg (2.3 𝑚/𝑠 )
The passenger sled pulls the cargo sled with a force 𝐹!" = 97 𝑁 in the +x-direction, so the cargo sled
pulls back with a force 𝐹!" = 97 𝑁 in the –x-direction.
SELF-EXPLANATIONPROMPTS
1. RephraseNewton’sSecondLawinyourownwords.
2. Describetheinteractionbetweenthedogteam,thesleds,theground,andgravity(i.e.,the
dogpullsthepassengersled….).
3. Explaineachcomponentofthefree-bodydiagramofthepassengersled.Bespecific.