Mechanics 8.2 - Newton's second law and resolving forces

Mechanics 8.2 - Newton's second law and resolving forces

Connect a force diagram to acceleration using Newton's second law. You will form signed equations along a straight line, work with perpendicular components and simple two-dimensional vectors, and choose useful axes for particles on smooth inclined planes.

Resultant force and acceleration

Can a particle moving right have an acceleration to the left? Start with a simpler force model: a 2 kg2\,\mathrm{kg} particle has forces 18 N18\,\mathrm N right and 5 N5\,\mathrm N left. The resultant is 13 N13\,\mathrm N right. Newton's second law tells us how that resultant changes the velocity; the particle's current motion does not decide the force direction.

You already know

Add forces as vectors. In one dimension, choose a positive direction and use signs. Zero resultant means constant velocity; a non-zero resultant changes velocity.

Mass measures how strongly the particle resists a given change in velocity. The same resultant produces half the acceleration when the mass is doubled.

Newton's second law

For a particle of constant mass mm,

∑F=ma.\sum\mathbf F=m\mathbf a.

Use mass in kg\mathrm{kg}, acceleration in m s−2\mathrm{m\,s^{-2}} and force in N\mathrm N. Here we use constant forces and constant acceleration. Not in the formula booklet: learn it.

The left side is the resultant of all forces on the particle. The units agree because 1 N=1 kg m s−21\,\mathrm N=1\,\mathrm{kg\,m\,s^{-2}}.

[DIAGRAM: asset_slug: p3_8_2_understand_and_use_newton_s_second_law_for_forces_in_straight_lines__diagram_01; asset_name: Mechanics 8.2 - Newton's second law and resolving forces - diagram 01; recommended_method: drawn_math; description: Draw a clean one-dimensional force diagram for a particle on a straight horizontal line. Show a positive direction arrow to the right, a rightward force of 18 N, a leftward force of 5 N, and rightward acceleration a. Include the equation 18 - 5 = 2a and the conclusion that the resultant force is 13 N to the right. Use spacious labels, real arrowheads, white background, and NovaLearn grey only.]
Diagram

Translate the diagram into one complete equation before using the calculator.

Worked example 1: resultant then acceleration

Question. The only horizontal forces on a 2 kg2\,\mathrm{kg} particle are 18 N18\,\mathrm N right and 5 N5\,\mathrm N left. Find its horizontal acceleration.

Plan. Take right as positive and divide the resultant by the mass.

18−5=2aNewton’s second law13=2aevaluate the resultanta=6.5 m s−2divide by mass\begin{aligned} 18-5&=2a &&\text{Newton's second law}\\ 13&=2a &&\text{evaluate the resultant}\\ a&=6.5\,\mathrm{m\,s^{-2}} &&\text{divide by mass} \end{aligned}

Why use 1313 rather than 1818? The leftward force also acts on the particle.

Answer. 6.5 m s−26.5\,\mathrm{m\,s^{-2}} to the right.

Check. ma=2(6.5)=13 Nma=2(6.5)=13\,\mathrm N, the signed resultant.

Change only the mass. This isolates why a force is not itself an acceleration.

Worked example 2: double the mass

Question. The same two forces act on a particle of mass 4 kg4\,\mathrm{kg}. Find its horizontal acceleration.

Plan. Keep the resultant and change the mass in Newton's second law.

18−5=4athe forces are unchangeda=134=3.25 m s−2divide by the new mass\begin{aligned} 18-5&=4a &&\text{the forces are unchanged}\\ a&=\frac{13}{4}=3.25\,\mathrm{m\,s^{-2}} &&\text{divide by the new mass} \end{aligned}

Why is the acceleration smaller? A larger mass needs a larger resultant for the same acceleration.

Answer. 3.25 m s−23.25\,\mathrm{m\,s^{-2}} to the right.

Check. 3.253.25 is half of 6.56.5, as the mass has doubled.

Keeping force and mass in their own units helps prevent an extra factor of gravity appearing.

Watch out

The equation is 18−5=2a18-5=2a, not 18=2a18=2a and not 18−5=2ga18-5=2ga. Mass is 2 kg2\,\mathrm{kg}; multiplying it by gg would turn it into a weight.

Use a fresh force pair and mass.

Scalar equations in one direction

Signs describe the chosen axis, not whether the particle is speeding up. A negative acceleration is opposite to your positive direction. It slows a particle moving positively, but speeds a particle moving negatively.

Equation of motion

An equation of motion for a chosen particle and direction sets the signed resultant force in that direction equal to mass times the signed acceleration in that same direction.

Write the direction first, then include every force component along it.

Worked example 3: a positive acceleration

Question. A 4 kg4\,\mathrm{kg} particle moves right along a horizontal line. A driving force of 30 N30\,\mathrm N acts right and a resistance of 6 N6\,\mathrm N acts left. Find its acceleration.

Plan. Take right as positive and subtract the resistance.

30−6=4aresolve along the line24=4aevaluate the resultanta=6 m s−2solve\begin{aligned} 30-6&=4a &&\text{resolve along the line}\\ 24&=4a &&\text{evaluate the resultant}\\ a&=6\,\mathrm{m\,s^{-2}} &&\text{solve} \end{aligned}

Why state the direction? The value 66 means right only because right was chosen positive.

Answer. The particle accelerates right at 6 m s−26\,\mathrm{m\,s^{-2}}.

Check. With left positive, 6−30=4a6-30=4a gives a=−6a=-6, the same rightward acceleration.

Change only the resistance from 6 N6\,\mathrm N to 38 N38\,\mathrm N. The particle is still initially moving right.

Worked example 4: moving right, accelerating left

Question. A 4 kg4\,\mathrm{kg} particle moves right. Its driving force is 30 N30\,\mathrm N right and its resistance is 38 N38\,\mathrm N left. Find its acceleration and describe its immediate motion.

Plan. Keep right positive, then interpret the acceleration relative to the velocity.

30−38=4aall forces along the linea=−2 m s−2divide −8 by 4\begin{aligned} 30-38&=4a &&\text{all forces along the line}\\ a&=-2\,\mathrm{m\,s^{-2}} &&\text{divide }-8\text{ by }4 \end{aligned}

Why does it slow down? Its rightward velocity and leftward acceleration have opposite directions.

Answer. It accelerates left at 2 m s−22\,\mathrm{m\,s^{-2}}, so it initially slows down.

Check. The resistance exceeds the drive by 8 N8\,\mathrm N, consistent with a leftward resultant.

This answers our opening question: motion and acceleration can point in opposite directions. Keep the model only for the interval in which the force directions given still apply; resistance may reverse if motion reverses.

Exam tip

Show the full equation of motion. A negative acceleration needs a direction or a clear positive-axis statement; it is a physical conclusion, not an algebra error to remove.

Decide both the acceleration and its effect on the current motion.

Perpendicular component equations

An angled pull can affect two equations at once. A 40 N40\,\mathrm N pull at 30∘30^\circ above a horizontal plane has rightward component 40cos⁡30∘40\cos30^\circ and upward component 40sin⁡30∘40\sin30^\circ. Only the first belongs in the horizontal equation.

You already know

For a force of magnitude PP at angle θ\theta above the horizontal, its components are Pcos⁡θP\cos\theta horizontally and Psin⁡θP\sin\theta vertically. Weight is mgmg vertically downwards.

If a particle stays on a horizontal plane, its vertical acceleration is zero even while it accelerates horizontally.

Component equations

For perpendicular axes,

∑Fx=max,∑Fy=may.\sum F_x=ma_x,\qquad \sum F_y=ma_y.

If the particle remains on a horizontal plane, ay=0a_y=0; a non-negative normal reaction must be consistent with contact. Not in the formula booklet: learn it.

Use g=9.8 m s−2g=9.8\,\mathrm{m\,s^{-2}} in the following examples. A smooth surface supplies no friction.

Worked example 5: pulling above the horizontal

Question. A 5 kg5\,\mathrm{kg} particle is pulled by 40 N40\,\mathrm N at 30∘30^\circ above a smooth horizontal plane. It remains on the plane. Find its horizontal acceleration and normal reaction.

Plan. Resolve horizontally for acceleration and vertically for reaction.

40cos⁡30∘=5ahorizontal equationa=43 m s−2use cos⁡30∘=3/2R+40sin⁡30∘−5g=0vertical acceleration is zeroR=49−20=29 Nsolve the vertical balance\begin{aligned} 40\cos30^\circ&=5a &&\text{horizontal equation}\\ a&=4\sqrt3\,\mathrm{m\,s^{-2}} &&\text{use }\cos30^\circ=\sqrt3/2\\ R+40\sin30^\circ-5g&=0 &&\text{vertical acceleration is zero}\\ R&=49-20=29\,\mathrm N &&\text{solve the vertical balance} \end{aligned}

Why use two equations? The pull has two components, but horizontal acceleration does not create vertical acceleration.

Answer. a=43 m s−2a=4\sqrt3\,\mathrm{m\,s^{-2}}, rightwards, and R=29 NR=29\,\mathrm N.

Check. The pull supports 20 N20\,\mathrm N of the 49 N49\,\mathrm N weight; R>0R>0, so contact is possible.

Now push at 30∘30^\circ below the horizontal instead. The horizontal component is unchanged, while the vertical component reverses.

Worked example 6: pushing below the horizontal

Question. A 5 kg5\,\mathrm{kg} particle is pushed by 40 N40\,\mathrm N at 30∘30^\circ below the horizontal, towards the right, on a smooth horizontal plane. Find its acceleration and normal reaction.

Plan. Reverse the sign of the vertical force component only.

40cos⁡30∘=5asame horizontal componenta=43 m s−2same horizontal accelerationR−40sin⁡30∘−5g=0push and weight both act downR=20+49=69 Nsolve for reaction\begin{aligned} 40\cos30^\circ&=5a &&\text{same horizontal component}\\ a&=4\sqrt3\,\mathrm{m\,s^{-2}} &&\text{same horizontal acceleration}\\ R-40\sin30^\circ-5g&=0 &&\text{push and weight both act down}\\ R&=20+49=69\,\mathrm N &&\text{solve for reaction} \end{aligned}

Why is the reaction larger? The plane must support both the weight and the downward push.

Answer. a=43 m s−2a=4\sqrt3\,\mathrm{m\,s^{-2}}, rightwards, and R=69 NR=69\,\mathrm N.

Check. The reaction increases by 40 N40\,\mathrm N, twice the reversed 20 N20\,\mathrm N vertical component.

That is a useful surprise: the same horizontal acceleration can accompany very different normal reactions. It follows exactly from the two independent component equations.

Watch out

Do not write 40−5g=5a40-5g=5a for horizontal motion: those forces point in different directions. Resolve first and use one equation per direction.

Keep the downward component in the vertical equation here.

Vector form in i and j

Vector notation packages two component equations into one line. If the resultant is (12i−6j) N(12\mathbf i-6\mathbf j)\,\mathrm N on a 3 kg3\,\mathrm{kg} particle, then 3a=12i−6j3\mathbf a=12\mathbf i-6\mathbf j, so a=(4i−2j) m s−2\mathbf a=(4\mathbf i-2\mathbf j)\,\mathrm{m\,s^{-2}}.

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