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I know that the formula for work, $W = FS\cos\theta$, where $F$ is the applied force, $S$ is the displacement of the object and $\theta$ is the angle between the applied force and the displacement of the object. When the object moves along the direction of the force, then we use $W = FS$. But why do we multiply it with $\cos\theta$ when the object doesn't move along the direction of the force?

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  • $\begingroup$ $\begin{aligned}W=\int \overrightarrow{F}\cdot \overrightarrow{ds}=\int \left( \overrightarrow{F}_{11}+\overrightarrow{F}_{\bot }\right) \cdot d\overrightarrow{s}= \int \overrightarrow{F}_{11}\cdot \overrightarrow{ds}=\int F\cos \left( \varphi \right) ds\end{aligned}$ $\endgroup$ – Eli Jun 9 at 18:56
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In general, work is a dot product between two vectors. You must integate over the path, $W=\int \vec F\cdot \mathrm d\vec s$, but if the force is constant, this simplifies to:

$$W=\vec F\cdot \vec s.$$

A dot product is mathematically the parallel components multiplied together. We could write it as

$$W=\vec F\cdot \vec s \quad\Leftrightarrow\quad W=F_\parallel \;s\quad\Leftrightarrow\quad W=F\;s_\parallel$$

if we wanted to. These three versions are all equivalent. The trick we need is how to get the parallel component only. If the forces are already parallel, then no biggie. Just multiply them without change:

$$W=F\;s\qquad \leftarrow \text{if parallel.}$$

If not, then it turns out that the cosine function can help. Remember from the unit-circle definition how cosine is defined: It is the horizontal distance while sine is the vertical distance. The cosine and sine values constistute the two legs (cathetusses) of a right-angled triangle in which the hypotenuse is 1. Scale it up to a ypotenuse with the length of your vector $F$ and the legs are now $F$ times longer than before, so $\cos(\theta)F$ and $\sin(\theta)F$.

These legs are the horizontal and vertical components, respectively, of the vector $F$, meaning:

$$F_\parallel=\cos(\theta)F\qquad\text{ and }\qquad F_\perp=\sin(\theta)F.$$

We could do the same for the vector $s$. In both cases, when we replace either $F_\parallel$ or $s_\parallel$ in work formula, we introduce this cosine term:

$$W=\vec F\cdot \vec s \quad\Leftrightarrow\quad W=F_\parallel \;s\quad\Leftrightarrow\quad W=F\;s_\parallel \quad\Leftrightarrow\quad W=Fs\cos(\theta).$$


The dot product uses cosine in this way. The cross product uses sine (used in the torque formula e.g.). To remember this technique for other uses, try to memorize that the cosine is equal to the adjacent leg of the right-angled triangle over the hypotenuse. And sine is equal to the opposite leg over the hypotenuse:

$$\cos(\theta)=\frac{\text{adjacent leg}}{\text{hypotenuse}}\qquad \text{ and }\qquad \sin(\theta)=\frac{\text{opposite leg}}{\text{hypotenuse}}.$$

(I don't know if there is some smart mnemonic in English - in Danish I typically say "cos er hos", because "hos" means "at / belongs to".) Since the hypotenuse always is the full vector length, then just flip it over to the over side of the equal sign, and then you have the expression you need for the parallel component (adjacent leg) or perpendicular component (opposite leg).

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Because the force applied on the object produces two effects - changing kinetic energy of the object by changing its speed and curving its path.

The work is supposed to tell us, how much kinetic energy was transferred to the object (or from it). The kinetic energy is however increased only by the component of the force in the direction of the motion, which is why the cosine appears. The perpendicular component serves only to curve the trajectory and there is no energy associated with the curvature of the motion so we are not interested in this component.

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  • $\begingroup$ THIS is the key, nice answer. I suggest you to include a circular motion example. In circular motion, parallel force increases kinetic energy while the normal component only maintains the rotation. That is to say, components parallel to motion increase velocity, while normal components only change trajectoy but not the modulus of the velocity $\endgroup$ – FGSUZ Jun 9 at 15:45
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Work done is a scalar quantity and it is defined by the product of two vectors Force and displacement. Now dot product of two vectors give a scalar quantity. So work done must be a dot product of the force and displacement vectors. That's why $W=FS\cos(\theta)$

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  • $\begingroup$ $|\vec{F}|\cdot|\vec{S}|$ is also a scalar quantity. The presence of the coseine has a deeper meaning than just making it a scalar $\endgroup$ – FGSUZ Jun 9 at 15:42
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To give and intuitive idea:

Imagine a person carrying a heavy suitcase fitted with wheels.

As long as the person pulls his luggage on a flat terrain, he does (ideally) no work: The (gravitational) force is perpendicular to the motion.

If the person arrives at a place when he/she has to pull his luggage say along an uphill ramp, then the more inclined the ramp, the more works said person will have to provide to go forward, because the larger the angle, the more the motion of the suitcase is aligned withe the force of gravity.

In this example the $cos(\alpha)$ measure how much the motion of the the suitcase is aligned withe gravity (i.e) with the vertical axis.

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$$W=\mathbf{F} \cdot \mathbf{S}$$ (work done is the dot product between the force applied and the displacement of the system)

There is another concept which is torque $$ \boldsymbol{\tau} = \mathbf{F} \times \mathbf{S}$$ (torque is the cross product or vector product of force applied and displacement of the system)

Dot product between the two vectors (both force and displacement are vector quantities here) is $\mathbf{F} \cdot \mathbf{S} = FS \cos\theta$ ($\theta$ is the angle between the two vectors). In conclusion, we measure how much force is applied in the direction of displacement or how much displacement happened in the direction of force.

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  • $\begingroup$ Welcome to Physics SE. First of all, try to lean MathJax to make your posts easier to read math.meta.stackexchange.com/questions/5020/…. Second, work in mechanics is measure of how much energy is transformed to/from kinetic energy. Measure of how much force is applied in the direction of displacement is tangential component of the force, not work. And third, I do not see question, nor even any mention of "work being done in some angle" so your answer seems to me mostly irrelevant. $\endgroup$ – Umaxo Jun 9 at 12:08

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