16.6 Resultant forces and dynamics in a plane — Learn

Prerequisites: resolving forces and coplanar equilibrium (Section 16.5); Newton's second law (Section 16.2). A friction or resistance force is given as data in this section; the model is Section 16.7.

By the end of this section you can: find the resultant of forces that are not at right angles; choose one axis along the acceleration; use Newton's second law in two directions; and find the acceleration of a body pulled at an angle or sliding on a slope.

1. The resultant of forces in a plane

Forces in a plane add by their components. Write each force as a column vector of its $x$ - and $y$ -components, then add the columns. The resultant has size

R = \sqrt{(\text{sum of } x)^2 + (\text{sum of } y)^2}

and a direction you can find with the tangent ratio. Only when the forces are all in one line, or at right angles, can you add their sizes directly.

2. Newton's second law in two directions

A body can accelerate along only one line at a time. Choose that line as one axis, so

\text{resultant along the motion} = ma,

and at right angles to the motion there is no acceleration, so

\text{resultant at right angles} = 0.

The second equation is usually what finds the normal reaction. The first is what finds the acceleration. Both are needed when a force acts at an angle.

Example 1. A force of $60$ N acts at $30^\circ$ above the horizontal and a force of $100$ N acts at $120^\circ$ . Find the size of the resultant and its direction from the positive $x$ -axis.

Take components.

x = 60\cos 30^\circ + 100\cos 120^\circ = 51.96 - 50.00 = 1.96 \text{ N},

y = 60\sin 30^\circ + 100\sin 120^\circ = 30.00 + 86.60 = 116.60 \text{ N}.

The size is

R = \sqrt{1.96^2 + 116.60^2} \approx 116.62 \text{ N}.

The direction is

\tan^{-1}\left(\frac{116.60}{1.96}\right) \approx 89.04^\circ

from the positive $x$ -axis, which is almost straight up.

Example 2. A crate of mass $45$ kg is pulled along a horizontal floor by a rope that makes $25^\circ$ with the horizontal. The tension is $200$ N and a resistance of $80$ N opposes the motion. Find the acceleration and the normal reaction. Take $g = 9.8$ m s $^{-2}$ .

The weight is

W = 45 \times 9.8 = 441 \text{ N}.

Resolve vertically. There is no vertical acceleration, so the vertical forces balance:

R + 200\sin 25^\circ = 441.

R = 441 - 84.52 = 356.48 \text{ N}.

The rope lifts the crate a little, so the floor pushes with less than the full weight.

Resolve horizontally. The horizontal component of the tension moves the crate:

200\cos 25^\circ = 181.26 \text{ N}.

The resultant along the floor is

181.26 - 80 = 101.26 \text{ N},

a = \frac{101.26}{45} \approx 2.25 \text{ m s}^{-2}.

Example 3. A block of mass $10$ kg slides down a slope of $30^\circ$ . A friction force of $17$ N acts up the slope. Find the acceleration. Take $g = 9.8$ m s $^{-2}$ .

The weight is $98$ N. Take one axis down the slope and one perpendicular to it.

The component of the weight down the slope is

98\sin 30^\circ = 49 \text{ N}.

The friction opposes the sliding, so it acts up the slope. Newton's second law along the slope gives

49 - 17 = 10a,

32 = 10a, \qquad a = 3.2 \text{ m s}^{-2}.

Example 4. A block of mass $6$ kg rests on a horizontal table. It is joined by a light inextensible string over a smooth pulley at the table edge to a hanging mass of $4$ kg. A friction force of $14.7$ N acts on the block. Find the acceleration. Take $g = 9.8$ m s $^{-2}$ .

The hanging weight is $4 \times 9.8 = 39.2$ N and drives the system. The resultant along the motion is

39.2 - 14.7 = 24.5 \text{ N},

on a total mass of $6 + 4 = 10$ kg, so

a = \frac{24.5}{10} = 2.45 \text{ m s}^{-2}.

3. Try it: a crate pulled at an angle

The sliders set the tension in the rope, the angle of the rope, the mass of the crate and the resistance. The figure draws the pull of the rope and its two components, together with the weight and the normal reaction.

Predict first. The rope lifts the crate a little, so the floor pushes with less than the full weight. Before you move a control, predict what happens to the acceleration as you raise the rope from $0^\circ$ towards $60^\circ$ . Write down your prediction and your reason.

Test. Move the angle slider and watch the forward component, the normal reaction and the acceleration. Then set the angle to $0^\circ$ and compare the normal reaction with the weight.

Explain. Write two sentences. The first says why the normal reaction falls as the rope rises. The second says why that smaller normal reaction leaves the acceleration unchanged in this model.

Watch out. Three mistakes are common.

Writing $R = mg$ when a force has a vertical component. In Example 2 the rope's vertical component is $84.52$ N upwards, so $R = 356.48$ N, not $441$ N. Always resolve vertically before you write down the normal reaction.

Putting friction the wrong way. Friction always opposes sliding. In Example 3 the block slides down the slope, so friction acts up the slope. Re-check: the acceleration must come out in the direction the body actually moves.

Using the wrong sign for a downhill component. On the slope, the weight component $mg\sin\theta$ acts down the slope. If the motion is up the slope, that component is a negative term in the equation along the slope.

You should now be able to

add forces in a plane by their components;
find the size and direction of a resultant that is not along an axis;
choose one axis along the acceleration and one perpendicular to it;
write two equations, one for the motion and one for the balance;
find the normal reaction when an applied force has a vertical component;
find the acceleration of a body pulled at an angle or sliding on a slope;
solve a connected-particles problem when a resistance acts on one body.