# Two dimensional forces

### Two dimensional forces

#### Lessons

In this lesson, we will learn:

• How to solve force problems when force is applied at an angle
• How to solve force problems with inclines

Notes:

• When looking at forces in two dimensions, a force can point along the x or y axis, or at any angle in between. The net force acting on an object is found by adding all the forces acting on that object using vector addition.
• When solving for net force it can be helpful to break angled forces into x and y components so that the forces is the x and y directions can be added separately.
• When an object is on a slope, it tends to be pulled down the slope by gravity. We can understand why gravity pulls the object down the slope if we break the force of gravity into two components: one that is parallel to the slope, and one that is perpendicular.
• We can redefine the x direction to be parallel to the slope and the y direction to be perpendicular to the slope for a particular problem. Essentially, we "tilt" the axes to line up with the slope. The components can then be solved like normal x and y components.
• $\vec{F}_{x}$ represents the amount of $\vec{F}_{g}$ which is pulling the object down the slope.
• $\vec{F}_{y}$ represents the amount of $\vec{F}_{g}$ pushing into the slope. It is balanced by the normal force from the slope pushing back on the box.
Newton's Second Law

$\Sigma \vec{F} = \vec{F}_{net} = m\vec{a}$

$\Sigma \vec{F}:$ sum of all forces, in newtons (N)

$\vec{F}_{net}:$ net force, in newtons (N)

$m:$ mass, in kilograms (kg)

$\vec{a}:$ acceleration, in meters per second squared $(m/s^{2})$

x and y Components of Force

$\vec{F}_{x or y} = \vec{F}\sin(\theta)$ (For the component opposite to $\theta$)

$\vec{F}_{x or y} = \vec{F}\cos(\theta)$ (For the component adjacent to $\theta$)

• Introduction
Introduction to two dimensional forces
a)
How to solve force problems when force is applied at an angle

b)
How to solve force problems with inclines

• 1.
Forces applied at an angle

A 4.25 kg box is pushed across the floor with a force of 31.0 N at 35.0° below the horizontal. The coefficient of friction between the box and the floor is $\mu$ = 0.125.

i. What is the normal force acting on the box?

ii. What is the force of friction acting on the box?

iii. Find the acceleration of the box.

• 2.
Forces on an incline
a)
A 3.70 kg box is placed at the top of an icy slope with a 28.0° incline. Assume no friction. Find the acceleration of the box down the slope.

b)
A 347.5 kg container is pulled up a 38.0° by a winch at 1.00 m/s. The coefficient of friction between the ramp and the box is 0.234. How much force does the winch need to pull with to keep the container moving up the ramp with this speed?