A 20.0-kg crate sits at rest at the bottom of a 11.0 m -long ramp that is inclined at 36 ∘ above the horizontal. A constant horizontal force of 290 N is applied to the crate to push it up the ramp. While the crate is moving, the ramp exerts a constant frictional force on it that has magnitude 65.0 N.
The relevant forces in the direction of displacement:
Fx = Fcos(θ)-friction
Fy = mgsin(θ)
Putting it all together:
Work = (Fcos(θ) - friction - mgsin(θ))(11)
I know which forces to prioritize, but what I don't quite understand is why I have to subtract Fx-Fy when I plug in F to the Work equation.
Any clarification would be greatly appreciated.
Fx = Fcos(θ)-friction friction is not in the x direction. It is down the plane.
Work=force*distance (both vectors, the dot product). I wlll do this as dot products, with i and j vectors
= 65i(cosTheta(11icosTheta+11jsinTheta)+ -forcefriction(-i)*11iCostheta
well the second term -i*i is -1; the first tem i*j has a dot product of 1, and the second i*j is zero. you put the -force friction because you are pushing against it. force opposing friction=-forcefriction
so finally,
work= 65*11*cos2Theta + forcefriction*11CosTheta where the force opposing friction is 65*cosTheta
In this scenario, the horizontal force (Fx) and the vertical force (Fy) are acting at different angles with respect to the displacement of the crate along the ramp. The work done by a force is equal to the dot product of that force and the displacement vector. The dot product is calculated as the product of the magnitudes of the two vectors multiplied by the cosine of the angle between them.
In this case, the force (F) you need to consider is the resultant force, which is the vector sum of Fx and Fy. The magnitude of the resultant force is given by:
|F| = √(Fx² + Fy²)
So, when you are calculating the work done by the force (F) on the crate, you need to use the magnitude of the resultant force |F|, which is equal to √(Fx² + Fy²), in the Work equation:
Work = (√(Fx² + Fy²) - friction - mgsin(θ))(11)
Now, if you expand the equation further, you will see that Fx and Fy are involved separately:
Work = (√(Fx² + Fy²) - friction - mgsin(θ))(11)
= (Fx - friction - mgsin(θ))(11)
So, you are subtracting Fx (the horizontal component of the force) and Fy (the vertical component of the force) when you plug in F to get the correct magnitude of the resultant force in the Work equation.
By considering both Fx and Fy separately, you take into account the effect of both the horizontal and vertical components of the force, relative to the direction of displacement. This approach allows you to accurately calculate the work done on the crate taking into account all the relevant forces involved.