Nice, one thing I have problem understanding is, at 1:08 why is it that you assumed the angle between Fx and Fy is 90 degrees? If I had drawn the Fp vector a bit longer and connected it with Fx, wouldn't that have made an angle with Fp and Fy a 90 degree angle? I mean what specifically made you choose that angle as the 90° and labeled the others after that?
For this problem it is most convenient to set the x-axis as the sloped part of the ramp. This makes the y-axis perpendicular to the x-axis (perpendicular to the ramp’s surface). For the forces involved in this problem, trigonometry is used to split the forces into their components in these new x and y axis.
If you are talking about what grade as in high school levels, this is usually either a senior level high school class (12th grade) or a freshman level college class.
That’s only F net in the “y” direction which in this problem is perpendicular to the surface of the ramp. All this means is that the box doesn’t fly off of the ramp or sink down into the ramp. So, the y components of the force due to gravity, normal force, and vertical pulling force will cancel out because there is no acceleration in the y direction. There is still a force on the box in the “x” direction.
Work is equal to the dot product of Force and displacement. The definition of the dot product states that work is equal to the force times the displacement, times the angle between the force and displacement. In this problem, delta X is up the ramp and the force of friction is down the ramp. The angle between the displacement and frictional force is 180 degrees.
Nice, one thing I have problem understanding is, at 1:08 why is it that you assumed the angle between Fx and Fy is 90 degrees? If I had drawn the Fp vector a bit longer and connected it with Fx, wouldn't that have made an angle with Fp and Fy a 90 degree angle? I mean what specifically made you choose that angle as the 90° and labeled the others after that?
For this problem it is most convenient to set the x-axis as the sloped part of the ramp. This makes the y-axis perpendicular to the x-axis (perpendicular to the ramp’s surface). For the forces involved in this problem, trigonometry is used to split the forces into their components in these new x and y axis.
@@ZeroWaitTutoring Alright, thanks!
What grade is this
If you are talking about what grade as in high school levels, this is usually either a senior level high school class (12th grade) or a freshman level college class.
@ZeroWaitTutoring OK thanks 😊
Why is F net equal to 0 tho? 🤔
That’s only F net in the “y” direction which in this problem is perpendicular to the surface of the ramp. All this means is that the box doesn’t fly off of the ramp or sink down into the ramp. So, the y components of the force due to gravity, normal force, and vertical pulling force will cancel out because there is no acceleration in the y direction.
There is still a force on the box in the “x” direction.
Where from the cos 180
Work is equal to the dot product of Force and displacement. The definition of the dot product states that work is equal to the force times the displacement, times the angle between the force and displacement. In this problem, delta X is up the ramp and the force of friction is down the ramp. The angle between the displacement and frictional force is 180 degrees.