R2, R5 - Resultant forces and Newton's second law

R2, R5 - Resultant forces and Newton's second law

Add forces as vectors and use their resultant to determine acceleration (R2, R5). You will connect constant forces to familiar motion equations, then use calculus and differential equations when a constant-mass particle experiences a force that varies with time or speed.

Adding force components

Can the same driving force and resistance law speed one particle up while slowing another down? Suppose a drive is 12 N12\,\mathrm N right and resistance has magnitude 6v N6v\,\mathrm N left, where vv is rightward speed. At v=1v=1 the drive wins; at v=3v=3 resistance wins. To predict either motion, first find the resultant rather than choosing one force.

You already know

A force is a vector. Add like signed components, and use Pythagoras for a vector's magnitude. Zero resultant gives constant velocity; a nonzero resultant changes velocity.

For a simple horizontal pair, 12 N12\,\mathrm N right and 6 N6\,\mathrm N left give 6 N6\,\mathrm N right, not 18 N18\,\mathrm N. With two perpendicular directions, carry out that signed addition separately in each direction.

Resultant force

The resultant force is the vector sum of all forces acting on the chosen body. It has the same effect on the particle's acceleration as those forces combined.

A vector-addition diagram may translate arrows head-to-tail. Its resultant joins the first tail to the last head. This is different from a free-body diagram, where the arrows identify actual interactions on one body.

Resultant in components

For perpendicular unit vectors i\mathbf i and j\mathbf j,

F=Fxi+Fyj∣F∣=Fx2+Fy2\mathbf F=F_x\mathbf i+F_y\mathbf j \qquad |\mathbf F|=\sqrt{F_x^2+F_y^2}

Find FxF_x and FyF_y by adding the signed components of all the forces. Direction needs a reference axis and the correct quadrant; for a nonzero resultant, use the component triangle.

Not in the formula booklet: learn it.

If both components are zero, the resultant has zero magnitude and no direction. Otherwise, a calculator's tan⁡−1(Fy/Fx)\tan^{-1}(F_y/F_x) alone may select the wrong quadrant.

Worked example 1: head-to-tail addition

Question. Forces (8i+3j) N(8\mathbf i+3\mathbf j)\,\mathrm N and (−2i+5j) N(-2\mathbf i+5\mathbf j)\,\mathrm N act on a particle. Find their resultant, its magnitude and its direction anticlockwise from i\mathbf i, to one decimal place.

Plan. Add components, use Pythagoras, then choose the angle from the signs.

Head-to-tail addition gives the resultant 6i + 8j N

F=(8−2)i+(3+5)jadd like components=6i+8j∣F∣=62+82=10 NPythagorasθ=tan⁡−1(86)=53.130…∘quadrant I\begin{aligned} \mathbf F &= (8-2)\mathbf i+(3+5)\mathbf j && \text{add like components} \\ &=6\mathbf i+8\mathbf j \\ |\mathbf F| &= \sqrt{6^2+8^2}=10\,\mathrm N && \text{Pythagoras} \\ \theta &= \tan^{-1}\left(\frac86\right)=53.130\ldots^\circ && \text{quadrant I} \end{aligned}

Why this step? The positive components place the resultant above and to the right, so the acute angle is measured from the positive i\mathbf i direction.

Answer. The resultant is (6i+8j) N(6\mathbf i+8\mathbf j)\,\mathrm N, with magnitude 10 N10\,\mathrm N and direction 53.1∘53.1^\circ anticlockwise from i\mathbf i.

Check. 10cos⁡θ=610\cos\theta=6 and 10sin⁡θ=810\sin\theta=8. The head-to-tail endpoint is (6,8)(6,8).

Worked example 2 reflects the force arrangement in the vertical axis: only the signs of horizontal components change. The magnitude stays the same, but the direction needs a quadrant correction.

Worked example 2: a resultant in quadrant II

Question. Forces (−8i+3j) N(-8\mathbf i+3\mathbf j)\,\mathrm N and (2i+5j) N(2\mathbf i+5\mathbf j)\,\mathrm N act on a particle. Find their resultant, magnitude and direction anticlockwise from i\mathbf i, to one decimal place.

Plan. Add signed components, then measure the quadrant-II angle from the positive horizontal axis.

F=(−8+2)i+(3+5)j=−6i+8jsum∣F∣=(−6)2+82=10 Nsquares remove signsϕ=tan⁡−1(86)=53.130…∘reference triangleθ=180∘−ϕ=126.869…∘quadrant II\begin{aligned} \mathbf F &=(-8+2)\mathbf i+(3+5)\mathbf j=-6\mathbf i+8\mathbf j && \text{sum} \\ |\mathbf F| &=\sqrt{(-6)^2+8^2}=10\,\mathrm N && \text{squares remove signs} \\ \phi &=\tan^{-1}\left(\frac86\right)=53.130\ldots^\circ && \text{reference triangle} \\ \theta &=180^\circ-\phi=126.869\ldots^\circ && \text{quadrant II} \end{aligned}

Why this step? The horizontal component is negative and the vertical component positive. The resultant points up-left, not down-right.

Answer. The resultant is (−6i+8j) N(-6\mathbf i+8\mathbf j)\,\mathrm N, with magnitude 10 N10\,\mathrm N and direction 126.9∘126.9^\circ.

Check. 10cos⁡θ=−610\cos\theta=-6, while 10sin⁡θ=810\sin\theta=8. Reflection changed the horizontal sign and preserved the length.

The addition method works for any number of forces, not just two. Equal and opposite vectors such as (3i−4j) N(3\mathbf i-4\mathbf j)\,\mathrm N and (−3i+4j) N(-3\mathbf i+4\mathbf j)\,\mathrm N cancel even though each has magnitude 5 N5\,\mathrm N.

Watch out

Wrong: ∣F1+F2∣=∣F1∣+∣F2∣|\mathbf F_1+\mathbf F_2|=|\mathbf F_1|+|\mathbf F_2| for every pair. Adding magnitudes discards direction. Right: add signed components first, then find the magnitude of their sum. Opposite 5 N5\,\mathrm N forces have zero resultant, not 10 N10\,\mathrm N.

Change the force components once more and check the quadrant before reporting the angle.

Second law in components

The resultant determines acceleration, not velocity. A rightward-moving particle can have a leftward resultant and slow down. Its motion at that instant does not change the signs of the forces.

Newton's second law

For a particle of constant mass mm,

F=maFx=maxFy=may\mathbf F=m\mathbf a \qquad F_x=ma_x \qquad F_y=ma_y

Here F\mathbf F is the resultant force and a\mathbf a is acceleration. One newton is 1 kg m s−21\,\mathrm{kg\,m\,s^{-2}}.

Not in the formula booklet: learn it.

Choose positive directions before writing equations. Reversing an axis changes both the force and acceleration components, leaving the physical prediction unchanged. For positive mass, acceleration has the direction of the resultant.

When the resultant is constant, acceleration is constant too. You may then use v=u+atv=u+at, which is in the formula booklet. The law that gives aa still has to be recalled.

Worked example 3: a retarding resultant

Question. A 6 kg6\,\mathrm{kg} particle initially moves right at 9 m s−19\,\mathrm{m\,s^{-1}}. Constant forces of 10 N10\,\mathrm N right and 22 N22\,\mathrm N left are its only horizontal forces. Find its acceleration and its velocity after 3 s3\,\mathrm s.

Plan. Take right positive, find the signed resultant, then use constant-acceleration motion.

10−22=6aresultant=maa=−2 m s−2divide by massv=9+(−2)(3)=3 m s−1v=u+at\begin{aligned} 10-22 &=6a && \text{resultant}=ma \\ a &=-2\,\mathrm{m\,s^{-2}} && \text{divide by mass} \\ v &=9+(-2)(3)=3\,\mathrm{m\,s^{-1}} && v=u+at \end{aligned}

Why this step? The force on the left is larger, so acceleration is leftwards even while the particle moves right.

Answer. Acceleration is 2 m s−22\,\mathrm{m\,s^{-2}} left. After 3 s3\,\mathrm s, velocity is 3 m s−13\,\mathrm{m\,s^{-1}} right.

Check. The velocity decreases by 6 m s−16\,\mathrm{m\,s^{-1}} in three seconds. Taking left positive would give a=2a=2, u=−9u=-9 and v=−3v=-3, describing the same motion.

Worked example 4 changes only the initial velocity. The force equation is unchanged, but the particle now reverses before the requested time.

Worked example 4: a reversal with unchanged forces

Question. The same 6 kg6\,\mathrm{kg} particle has the same constant forces, 10 N10\,\mathrm N right and 22 N22\,\mathrm N left, but initially moves right at 3 m s−13\,\mathrm{m\,s^{-1}}. Find its velocity after 3 s3\,\mathrm s and when it first comes to rest.

Plan. Keep a=−2 m s−2a=-2\,\mathrm{m\,s^{-2}}, then use signed velocity to locate the stop and subsequent direction.

v=3−2tright is positivev(3)=3−6=−3 m s−1requested time0=3−2tinstant at restt=1.5 s\begin{aligned} v &=3-2t && \text{right is positive} \\ v(3) &=3-6=-3\,\mathrm{m\,s^{-1}} && \text{requested time} \\ 0 &=3-2t && \text{instant at rest} \\ t &=1.5\,\mathrm s \end{aligned}

Why this step? A negative velocity now means motion leftwards. It is different from the negative acceleration, which acted throughout.

Answer. The particle first rests at 1.5 s1.5\,\mathrm s, then moves left; after 3 s3\,\mathrm s it moves left at 3 m s−13\,\mathrm{m\,s^{-1}}.

Check. At the stopping instant the resultant is still 12 N12\,\mathrm N left, so the particle does not remain at rest. The forces here are specified constant forces, rather than resistance that changes direction after a reversal.

In a plane, use the same law in each perpendicular direction. For example, divide each component of (−5i+10j) N(-5\mathbf i+10\mathbf j)\,\mathrm N by the same mass; do not divide its magnitude and then attach the original vector.

Watch out

Wrong: 22=6a22=6a, using just the larger force. Right: 10−22=6a10-22=6a when right is positive. Newton's second law uses the resultant of all relevant forces, with signs.

The next particle has a resultant already supplied as a vector, so there is no extra force sum to construct.

Time-dependent force and motion

A changing force can produce changing acceleration. If its i\mathbf i component is proportional to time, that component of acceleration is not constant over an interval. Integrate it rather than inserting one value into a constant-acceleration formula.

You already know

Acceleration is the derivative of velocity; displacement over an interval is the integral of velocity. Integrate a vector component by component, and use the initial velocity to determine each integration constant.

A constant mass lets you divide the entire force function by mm. The time variable remains in the acceleration function.

Force and calculus

For constant mass,

mdvdt=F(t)r(t)−r(t0)=∫t0tv(τ) dτm\frac{d\mathbf v}{dt}=\mathbf F(t) \qquad \mathbf r(t)-\mathbf r(t_0)=\int_{t_0}^{t}\mathbf v(\tau)\,d\tau

Here τ\tau is a dummy time variable. Both component constants must satisfy the initial condition.

Not in the formula booklet: learn it.

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