1. Waterballoon-face collision
In Armageddon, a NASA guy comments that a plan to shoot a laser at the asteroid is like “shooting a b.b. gun at a freight train.” What would it take to stop an out-of- control freight train using only b.b. guns?

2. Torque and equilibrium
A torque (acting alone) would make an object at rest rotate
Equilibrium:
SF=
St=
No acceleration of center of mass!
No angular acceleration either!

3. Torque and equilibrium
St=

4. Forces that cause torques
The line of force must not pass through the rotation axis
OR, equivalently…
A force must have a component which is perpendicular to the line between the axis and the force application point

5. Calculating torque
moment or lever arm or “r”: perpendicular (smallest) distance between the line of force and the axis of rotation.

6. Calculating torque
P1. Which force causes the largest torque about the left end?
A. The left 80 N force
B. The right 80 N force
You can answer this by calculating, or carefully extending the lines to draw the moment arms

7. Equilibrium
What are the support forces on the beam?

8. Equilibrium and torque hints
Draw a free-body-diagram of the object alone.
Include all forces in the diagram
The weight of objects acts at their center of gravity.
Decide on an “axis of rotation” to calculate torques, which can be chosen anywhere. (Best to choose axis where you don’t need to know forces)
Decide on + and – directions of rotation.
Find moment arms or perpendicular components of forces.
Use St = 0 and possibly SF=0 until you find all the information you need.
Center of gravity (or mass): The point at which gravity effectively acts, where all the weight can be considered to be concentrated.

9. What forces on the door are needed at the hinges to balance the door weighing 400 N? Assume that the top hinge bears all the weight. The distance between hinges is 1.2 m, and the door width is 1.0 m

11. Find the tension in the horizontal rope. The boom weighs 400 N

12. Find the tension in the horizontal rope. The boom weighs 400 N
Find the forces on the pivot point

13. P2. The horizontal (x) force component of the pivot on the beam is
(hint: sum forces)
to the right
to the left
zero
P3. The vertical (y) force component of the pivot on the beam is
(hint: think torques using a different pivot point!) up
down

14. A ladder leans against a frictionless wall. The ground has friction
A ladder leans against a frictionless wall. The ground has friction. The ladder doesn’t slip
Draw a FBD of the ladder. You should have four forces.
P4. The ground’s frictional force is ________ compared to the wall’s normal force.
more than
less than
the same
P5. If the ground contact point is chosen as the axis, The torque of the wall’s normal force is ________ compared to the torque of the weight.