M1 4.1, M1 5.1 - Newton's laws and force components

M1 4.1, M1 5.1 - Newton's laws and force components

Connect force diagrams to acceleration using Newton's laws in M1 4.1. Resolve forces into perpendicular components using M1 5.1, then build scalar and vector equations for the chosen object.

Force and the three laws

A person pushes a trolley, and the trolley pushes back equally. Why can the trolley accelerate? The two forces act on different objects, so they do not cancel in the trolley's force equation.

A force is an interaction that can change motion. It has magnitude and direction; its SI unit is the newton, N\mathrm N. A force diagram includes only forces acting on one chosen object. The resultant force is their vector sum.

Newton's laws of motion

First law: a particle remains at rest or moves with constant velocity when the resultant force on it is zero.

Second law: for constant mass, the resultant force equals mass multiplied by acceleration, R=ma\mathbf R=m\mathbf a.

Third law: when one body exerts a force on another, the second exerts an equal-magnitude, opposite-direction force of the same type on the first. These forces act on different bodies.

One newton gives a mass of 1 kg1\,\mathrm{kg} acceleration 1 m s−21\,\mathrm{m\,s^{-2}}. Weight is the gravitational force on an object, with magnitude mgmg; mass itself is not a force. In this lesson, where required, use g=9.8 m s−2g=9.8\,\mathrm{m\,s^{-2}}.

Worked example 1: balanced forces on one body

Question. A trolley is pushed right by 10 N10\,\mathrm N and resisted left by 10 N10\,\mathrm N. Its vertical forces also balance. Explain what follows about its acceleration and whether it must be stationary.

Plan. Add forces on the trolley and apply the first and second laws.

R=10−10=0⇒a=0.R=10-10=0\quad\Rightarrow\quad a=0.

Answer. Acceleration is zero. It may be at rest or may move with constant velocity; the forces alone do not choose between these possibilities.

Check. Either behaviour has zero velocity change, consistent with zero acceleration.

Change only the body on which the second force acts. A third-law pair cannot be combined as two forces on the trolley.

Worked example 2: a third-law pair

Question. A hand pushes a trolley right with 10 N10\,\mathrm N. The trolley pushes the hand left with 10 N10\,\mathrm N. Do these two forces establish zero resultant force on the trolley?

Plan. Assign each force to its receiving body before adding anything.

The rightward force acts on the trolley; the leftward force acts on the hand. Only the first belongs on the trolley's diagram.

Answer. No. These are a third-law pair on different bodies. The trolley's acceleration depends on all the forces acting on the trolley.

Check. Adding the hand's force to the trolley's diagram would incorrectly mix two different systems.

The weight and normal reaction on a resting book are also not a third-law pair: both act on the book. The partner to the Earth's gravitational pull on the book is the book's gravitational pull on the Earth.

Resolving forces

A sloping force can be replaced by two perpendicular components whose vector sum is the original force.

Draw the right triangle beside the force. A component adjacent to the stated angle uses cosine; the opposite component uses sine. The choice follows the angle's location, so it is unsafe to memorise “horizontal always uses cosine”.

Resolving a force

For force magnitude FF making angle θ\theta above the positive horizontal,

Fx=Fcos⁡θ,Fy=Fsin⁡θ.F_x=F\cos\theta,\qquad F_y=F\sin\theta.

If the angle is measured from the positive vertical instead,

Fx=Fsin⁡θ,Fy=Fcos⁡θ.F_x=F\sin\theta,\qquad F_y=F\cos\theta.

Attach component signs from the diagram. These resolution formulae are not in the formula booklet: learn them.

[DIAGRAM: asset_name: The reference axis changes the components; asset_slug: m1-force-resolution; caption: Cosine belongs to the component adjacent to the given angle.; description: Two force triangles show a forty-newton force thirty degrees above horizontal, with horizontal component twenty root three and vertical component twenty, and a forty-newton force thirty degrees from vertical, with horizontal component twenty and vertical component twenty root three.]
Cosine belongs to the component adjacent to the given angle.

Worked example 3: angle from the horizontal

Question. A force of magnitude 40 N40\,\mathrm N acts 30∘30^\circ above the positive horizontal. Find its horizontal and vertical components.

Plan. Use cosine for the adjacent horizontal component and sine for the opposite vertical component.

Fx=40cos⁡30∘=203 N,Fy=40sin⁡30∘=20 N.F_x=40\cos30^\circ=20\sqrt3\,\mathrm N, \qquad F_y=40\sin30^\circ=20\,\mathrm N.

Answer. 203 N20\sqrt3\,\mathrm N right and 20 N20\,\mathrm N up.

Check. (203)2+202=40\sqrt{(20\sqrt3)^2+20^2}=40, recovering the original magnitude.

Change only the reference axis for the 30∘30^\circ angle. The component values swap.

Worked example 4: angle from the vertical

Question. A force of magnitude 40 N40\,\mathrm N acts upwards and right, making 30∘30^\circ with the positive vertical. Find its horizontal and vertical components.

Plan. The vertical component is now adjacent to the angle.

Fx=40sin⁡30∘=20 N,Fy=40cos⁡30∘=203 N.F_x=40\sin30^\circ=20\,\mathrm N, \qquad F_y=40\cos30^\circ=20\sqrt3\,\mathrm N.

Answer. 20 N20\,\mathrm N right and 203 N20\sqrt3\,\mathrm N up.

Check. Fx/Fy=1/3=tan⁡30∘F_x/F_y=1/\sqrt3=\tan30^\circ, as the angle from the vertical requires.

Components replace the original force in an equation; they are not extra forces to add on top of it.

Watch out

A 40 N40\,\mathrm N force does not become 40 N40\,\mathrm N horizontally and 40 N40\,\mathrm N vertically. Recombining its perpendicular components must recover magnitude 40 N40\,\mathrm N.

Read the reference axis before choosing sine or cosine.

Continue this lesson.

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