is work = F* S*cos theta is the total work?

Question

plz read the following very carefully.
we do work to move a box on an inclined plane...
we have to prove that work done = fs cos theta
for proving f s cos theta:
we can do by 2 methods,...

1)by finding the component of displacement of body in the direction of force.

here,...surely the work we are finding is the work we do in that particular direction?
so ...it is not equal to the full work ?

2)we find the component of force applied in the direction in which the object oves on the inclined plane.
as this is the direction in which the object is movng ,..surely this is the total work done.?

but in accord with the first method...we got ...F * S cos theta
in accord with the 2nd method,...we got ...Fcos theta* S
so both of them are equal...but how can that be when one is measuring only work along a specific direction and the other work in the whole?

i would really prefer an explanation in logic rather than by mathematics.
i know it will take time...but plz do help.

s = distance we moved the object on inclined palne.

i know work is scalar...but force is not.
actually,...the force needed to push up a inclined plane a box,...is a composite of two way forces...1 up and 2nd sideways...
this is what i'm saying...the component of the sideways one in the 1st method

2nd answerer-- yeah i understand that...but the composite force is given by the hypotenuse also...so if one force was acting in up and other side...then surely the hypotenuse will be bigger?
so how could it be less....yet i know that being cos theta value less than 90 degree,...it should decrease the original value...yet i dont somehow understand that how will the upward force decrease the side force(it has to if the final value is coming lower)? will it do that?

hey u,...orange...mind where u go...

Answer 1

hey catty,

You recognize that mathematically, the 2 expressions are equivalent, right?

If the force and the displacement were both in line, then
W = F*d.
The way I was taught this, if they are not in line
W = F*d*cos(theta) where theta is the angle between the 2. Period.

(OK, I know if you ask for a justification of the trig function, then you get into talk about components of the vectors [dislacement is a vector].)

So ... logically you see #2 as being the whole work. The component of the force that's along the incline does the work to move the box. The other component of the force is perpendicular to the first component (and to the resulting motion) and does no work. So that 2nd component is wasted force. It's still expended, but wasted.

Let's talk about #1 logically. The force is trying to push the box in a certain direction. But the incline only allows it to move along it's surface. So the box only moved s*cos(theta) in the direction the force wanted it to move. As far as the force is concerned, the component of the motion that deviated from the straight ahead direction is wasted motion.

So in either case, something is wasted.

Hope that helps,
sojsail

Answer 2

ok. I plan to artwork it without finding on the e book answer. The cos is a factor of the answer because of the fact in simple terms the area of stress for the time of action or replace action counts. keep in mind the question: How plenty artwork do you do on the container . in case you carry a one hundred kg container 50 meters, shifting the container horizontally? not likely a trick question,whether it confuses a lot of commencing up pupils. you're protecting the container up so your hands are making use of: Wt = MA = one hundred kg * 9.8 m/s^2 = 980 N. yet, that 980 N is utilized up and the container strikes horizontal. So, we could discover the area of the 980 N that is utilized in the horizontal course and that's: W = cos (perspective between stress and action) * stress * distance W = cos ninety * 980 * 50 W = 0 * 980 * 50 = 0 nevertheless you're protecting the container up, you do 0 artwork on the container because of the fact which you progression it horizontally. that's the comparable as though the container grow to be on a frictionless cart and you had to coach 0 stress horizontally to pass the cart 50 meters after a small push to get it shifting. Your hands are the comparable because of the fact the cart, all they do is carry the container up. not one of the stress is going to pass the container Now the topic: i assume the line is horizontal. If the line grow to be at an perspective then we would choose the cos term. If the line is horizontal, then the perspective is 0 and cos=one million and we are able to overlook approximately that element era. KE = artwork = stress * distance (that is authentic in simple terms because of the fact there is not any FRICTION) KE = six hundred N * one hundred m = 60,000 J and KE additionally = 0.5 MV^2 KE = 60,000 J = 0.5 *1000 kg * V^2 60,000 = 500 * V^2 V^2 = 60,000/500 = v^2 = one hundred twenty V = 10.95 m/s evidently the e book is authentic.

Answer 3

In the one case you get to apply the full force (F), but along a reduced distance (Scos(theta)). In the other case you apply a reduced force (Fcos(theta)) but along the full distance (S). That is why they are equal.

In answer to your additional question, the "upward" force is correlated to a decrease in the "side" force. One is proportional to cos(theta) the other is proportional to sin(theta). As one increases, the other decreases (within the 0-90 degree range).

Answer 4

F Cos

Answer 5

since work is a scalar, it does not depend on the direction.
But then again you cant take the component of a scalar(i.e. in this case "The distance travelled" is a scalar)
so u can take the component of force only,which is a vector

Answer 6

ok ! Let me clarify this,
Suppose a body has a displacement of s in a particular direction becuse of the force apllied to it, which do not have the same direction as s. Let this force be F.

We, resolve this displacement, in two rectangular components. S1 & S2 [say]

Work done in the direction of the force= F.S1. cos 0 deg = F*S1
But in case of S2 the displacement is perpendicular to F. So in case of displacement component S2, F remains as a no work force( angle between applied force & displacemrnt being 90 deg)

Hence the total work done = F.S1 + F.S2= F*S1+0=F*S1

Similarly, you may divide s in as many components as you like.
The S1,S2,.....,Sn
in that case if they make angles θ1,θ2,.....,θn with the direction of force.
If it makes an angle of θ with the main displacement.
Then, S1 cos θ1+S2 cos θ2+....+Sn cos θn = S cos θ
S1 sin θ1+S2 sin θ2+....+Sn sin θn = S sin θ
We choose the axes, parallel to the direction of the force.
Hence total work done=
F.S1 cos θ1+F.S2 cos θ2+....+F.Sn cos θn+F.S1 sin θ1+F.S2 sin θ2+....+F.Sn sin θn =F.S cosθ + F.S sinθ
Again, S sin θ is normal to the direction of applied force. So total work done in this direction = F*S sin θ * cos 90 deg. = 0

So, total work done, W = F. S cos θ( mind it is the dot product & cos θ is still on) = F*S cos θ *cos 0 deg ( S cos θ is in the direction of F)
= F*S cos θ.



The actual difficulty arises when we consider dissipative forces like friction, their direction is depedent on direction of displacement, so we have to take the component of force in direction of displacement for the sake of simplicity!!!


Now what does that remark mean?
oh! sorry! typing mistake!

Answer 7

no
the total work is
F*S cos theta+F*S sin theta = total work
but
F*S cos theta= only work on the horizontal
I learnt it as cos for the capitate
and sin for horizontal
i hope it is clear