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Forces and Angles
Physics 11
More bad jokes…
Review: Free-body diagrams:

Free-body diagrams are
used to show the relative
magnitude and direction of
all forces acting on an
object.

What if some of the forces
were not in the x or y
plane???????
Forces at angles:

We will break
the forces with
angles into their
x and y
components
using
trigonometry!
Review: Fnet

Remember we are only worried about Fnet in the
direction of the motion
 So
sometimes we really only need either the x or y
component of the force at an angle.
Example 1:

You need to move a 25.5
kg wooden crate across a
wooden floor (µ=0.20).
You pull with a rope
attached at a 15º to the
horizontal with a force of
276N.
a.
b.
Calculate the Fapp(X) and
Fapp(Y).
What is the acceleration
of the crate?
a.
Fapp(x)= 276 N * cos 15
= (276)(0.9659)
= 267 N
Fapp(y)= 276 N * sin 15
= (276)(0.2588)
= 71.4 N
=15º
b.




FBD
Only need to consider x axis forces as that is the plane of the motion.
Ff = µFN
= (0.20)[(25.5)(9.81)
= 50.0 N
Fnet= Fapp(x) + Ff
= 267N – 50N
= 217 N
Fnet= ma
217= (25.5)a
8.5 m/s2 = a
Is it easier to push or pull a sled?
Pulling a Sled

A child of 35kg is sitting on a sled that has a
coefficient of kinetic friction of 0.10 with the snow.
If the parent who is pulling the sled pulls up with an
angle of 25°, with what are the component forces?

Draw FBD
FBD
FN
Fa
Ff
Fg
Break Applied Force into
Components
Fax  Fa cos 
Fa
Fay
Fax
Fax  Fa cos( 25)
Fax  (.906) Fa
Fay  Fa sin 
Fay  Fa sin( 25)
Fay  (.423) Fa
Example: Pushing a Sled

A child of 35kg is on a sled that has a µk =0.10
with the snow. If the parent who is pushing the sled
pushes down with an angle of 25°, with what are
the component forces.

Draw FBD.
FBD
FN
Fa
Ff
Fg
Break Applied Force into
Components
Fax  Fa cos 
Fax  Fa cos( 25)
Fax
Fay
Fa
Fax  (.906) Fa
Fay  Fa sin 
Fay  Fa sin( 25)
Fay  (.423) Fa
Practice:

Force Components: Vectors WS
 Do


5d and 6c together
Page 171 #11, 12 (assume no Ff)
Page 208 # 24, 25