anonymous
  • anonymous
Calculate the Applied Force that will make the object move up the ramp with constant velocity. Because the ramp is frictionless, the only force pushing the object down the ramp is the component of gravity that is parallel to the ramp. This is equal to mg sin theta (θ) where m is the mass of the object, g is the acceleration due to gravity, and θ is the angle of incline. tthe mass is 100, the gravity is of course 9.8, and the inclined ramp is 10 degrees. if anything else is needed just ask but plese explain this procedure
Physics
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
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anonymous
  • anonymous
@YumYum247
anonymous
  • anonymous
I think this problem could be easily done by inspection, especially with the information that they've given you. But it's always best to break it down, which I will do for you!
anonymous
  • anonymous
I always love these force problems X)

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anonymous
  • anonymous
Here is the scenario that we have: |dw:1450342023766:dw|
anonymous
  • anonymous
Please note that there are other forces being applied, such as the normal force. However, as you will see, we will not need these. The forces that lie along this particular plane are the only ones that necessarily matter.
anonymous
  • anonymous
Now that we've drawn these forces, we should consider using one of the biggest laws: `Newton's 2nd Law` \[\huge \text{F}=ma\] However, in this case, we need to use a different "version" of this equation to find the net force, which is given by:\[\huge \sum \vec{\text{F}}=\sum_i^nm_ia_i\]Please note that we can only use this particular equation for the net force on the SAME plane. For example, we can only use the sum of forces in the x-direction. But we cannot sum x-directional forces and y-directional forces together. This also works for any u-v planes (aka, non-traditional x-y planes) such as inclined planes (our scenario here).
anonymous
  • anonymous
If you'd like, we can think of the inclined plane as the "x-axis" and analyze the forces as such. We can see that our equation becomes\[\huge \text{F}_\text{applied}-\text{F}_\text{g}=ma\]
anonymous
  • anonymous
However, our equation *actually* becomes \[\huge \text{F}_\text{applied}-\text{F}_\text{g}=0\]Why? Because we're told that the object is moving at a constant velocity. This means that the acceleration is 0
anonymous
  • anonymous
So what is Fg? It is the component of the weight that is down the slope. As given in the problem, this particular component is:\[\huge \text{F}_\text{g}=mg \sin(\theta)\]
anonymous
  • anonymous
Plugging this into our equation from before, we see that \[\huge \text{F}_\text{applied}-mg \sin(\theta)=0\]Which of course means that\[\huge \text{F}_\text{applied}=mg \sin(\theta)\] This is the minimum force needed to keep the boxing moving upward a constant velocity
anonymous
  • anonymous
As always, I believe it important to have other scholars check work @Michele_Laino @IrishBoy123 @Astrophysics
Michele_Laino
  • Michele_Laino
that's right! @CShrix

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