# 9709. Mechanics 1. 2D Dynamics (Resolving Forces)

Horizontal and Vertical Components of Force

• Because force is a vector quantity, we can refer to a single force by means of two forces in perpendicular directions.  So a force F at angle 𝜽 to the horizontal can be expressed as a horizontal force FX and a vertical force FY.
• FX = Fcos𝜽;
• FY = Fsin𝜽.

If we have several forces acting in several directions, we can resolve the forces horizontally to find the net horizontal force.

We can also resolve them vertically to find the net vertical force.

For a particle to be in equilibrium (i.e. at rest or constant speed) the net horizontal force and net vertical force must both equal zero.

Worked Examples

• A girl pushes a toy with mass 2kg along the floor with a force of 5N. She applied the force at an angle of 35º to the horizontal. Neglecting resistance forces find the acceleration of the car;
• A box of mass 15kg is dragged along the floor at a constant speed of 1.2ms-1 by means of a rope at 30º to the horizontal. The tension from the rope is 50N. Calculate the frictional force resisting the montion and the normal contact force from the floor.
• A particle in equilibrium has 3 forces of magnitude 5N, 6N and FN acting on it in the directions shown. Find the values of F and 𝜽:

Exercise

Resolution in directions other than vertical and horizontal

In many situations it is more helpful to resolve in other directions, particularly parallel to the line along which a force is acting and perpendicular to that line.

If 𝜽 is the angle between the line of action of the force and the direction of the particle, then the component of the force in the direction of the particle is Fcos𝜽 and the component of the force perpendicular to the direction of the particle is Fsin𝜽.

Worked Examples

• A crate of mass 30kg is at rest on a ramp that is sloped at an angle of 18º to the horizontal. The crate is prevented from sliding down the ramp by friction. Find the frictional force and the normal contact force.
• A book of mass 5kg sits on a slope inclined at 20º to the horizontal. It is held in equilibrium and prevented from sliding down the slope by a frictional force. The book is then pulled up the slope at a constant speed, using a rope that makes an angle of 50º with the horizontal. Assume that the frictional force has the same magnitude as before and now acts down the slope. Find the tension in the rope.
• A rail track is laid on the floor of a quarry, sloping down at 11º to the horizontal from the side of the cliff. A truck is filled with stone and when full may have mass up to 600kg. It is prevented from running downhill by a cable from the truck to the top of the cliff, as shown below. For safety reasons the tension in the cable should not exceed 1600N. For the information above, find:
• The greatest angle that the cable should be allowed to make with the track; and
• The smallest value of the total normal contact force on the truck.

Exercise

Objects on slopes not in equilibrium

If the net force is non-zero, then the particle will accelerate according to NII.

We typically resolve forces parallel to the direction of motion.  For an object on a slope, this will be parallel to the slope.

We can also resolve forces perpendicular to the direction of motion (in which case acceleration will typically equal zero) in order to get a second equation to help find unknown forces (including the normal contact force).

Worked Examples

1. A raft of mass 50kg is pulled along a canal using two ropes. The rope to the left bank of the canal is at an angle of 20º to the direction of motion with a tension of 100N. The rope to the right bank of the canal is at an angle of 25º to the direction of motion. The resistance to the motion of the raft is 40N. Find the acceleration of the raft.
2. A cyclist starts from rest and moves along a straight, horizontal track. The forward driving force is a constant 72N and the resistances to forward motion are a constant 20N. The combined mass of the bicycle and the cyclist is 80kg
1. Calculate the acceleration of the cyclist;
2. How long does it take the cyclist to reach a speed of 8ms-1? How far does the cyclist travel in this time?
3. The cyclist starts to climb a hill when her speed reaches 8ms-1. The hill is at a constant angle of 10º to the horizontal. The driving force and resistances to motion are unchanged. Find the acceleration (or deceleration) of the cyclist on the hill.

Exercise

Finding magnitude and direction of the resultant force (and related acceleration)

By combining the effect of several forces in 2 dimensions, we can find a single resultant force.

This can be done by resolving all of the forces horizontally and vertically and then treating their horizontal sum as X and their vertical sum as Y.

The resultant force then has magnitude $\sqrt{X^2+Y^2}$ and direction $tan^-1(\frac{Y}{X})$ from the horizontal.

Don’t include the resultant force on a force diagram (I mean do a second diagram if you want to show resultant forces – don’t include it on the same diagram with the original forces).

Worked Examples

Exercise and General Exercises on this Topic