04+Dynamics+-+Newtons+Laws+of+Motion

toc =Handouts to look at:= Any of these have good information but the 4th talks about Atwood systems and tension forces.

Use www.physicsclassroom.com
Physics Classroom Handout that we did before: http://www.physicsclassroom.com/getattachment/curriculum/force2D/forces.pdf =Video:= []

Elevator Problems:
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Atwood Problems:
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2 Box Problem
the two box question on the review work (19) media type="custom" key="24661134"

Forces Review:
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Labs:
Lab 1 Lab 2 Lab 3

**Notes:**
Note: the problem where you have a cart going up an incline and another cart is released with velocity at the same time the first cart stops and both carts reach the bottom of the incline at the same time. (two part question-find the distance up the incline the first cart goes and the initial speed of the second cart) First of all you need to calculate the acceleration component acting along the incline. This is your gravitational field.

Now you can calculate how long for the first sled to go up and stop. (This will be the same time for it to go back down again.) And also how far up it goes.

Knowing that you know how long the second sled has to reach the bottom and hence what speed it must be released at.


 * Block on wedge block with horizontal force:**

Discussion
For problems that involve two blocks one on a table with friction the other suspended from a pulley, connected to the first block and hanging off the table. When using the Fnet=ma formula the m is the mass of block one plus the mass of block two. The acceleration and net force for block one is the same as the acceleration and net force for block two. ||
 * || [|DustinS.101]

Using Newton’s second law, the accelera- tion of the wedge of mass 18 kg and block of mass 5 kg is aAB = F mA + mB (1) Consider the free body diagram for block B m g sin  N mg aAB The condition that block B does not move <span style="display: block; height: 1px; left: -40px; overflow: hidden; position: absolute; top: 476.5px; width: 1px;">with respect to wedge A implies that its ac- <span style="display: block; height: 1px; left: -40px; overflow: hidden; position: absolute; top: 476.5px; width: 1px;">celeration down the wedge with respect to the <span style="display: block; height: 1px; left: -40px; overflow: hidden; position: absolute; top: 476.5px; width: 1px;">wedge is zero. Applying Newton’s second law <span style="display: block; height: 1px; left: -40px; overflow: hidden; position: absolute; top: 476.5px; width: 1px;">for B in the direction parallel and perpendic- <span style="display: block; height: 1px; left: -40px; overflow: hidden; position: absolute; top: 476.5px; width: 1px;">ular to the wedge yields <span style="display: block; height: 1px; left: -40px; overflow: hidden; position: absolute; top: 476.5px; width: 1px;">XF? : N − mB g cos  = mB aAB sin (1) <span style="display: block; height: 1px; left: -40px; overflow: hidden; position: absolute; top: 476.5px; width: 1px;">XFk : mB g sin  = mB aAB cos  (2) <span style="display: block; height: 1px; left: -40px; overflow: hidden; position: absolute; top: 476.5px; width: 1px;">Using Eq. 1 to solve for the normal force