×

# Torque

While calculating torque, we assume that the force to the moment is perpendicular to it. I was wondering, what would happen if its not perpendicular. Please help!

Note by Tanay Roman
4 years, 9 months ago

MarkdownAppears as
*italics* or _italics_ italics
**bold** or __bold__ bold
- bulleted- list
• bulleted
• list
1. numbered2. list
1. numbered
2. list
Note: you must add a full line of space before and after lists for them to show up correctly
paragraph 1paragraph 2

paragraph 1

paragraph 2

[example link](https://brilliant.org)example link
> This is a quote
This is a quote
    # I indented these lines
# 4 spaces, and now they show
# up as a code block.

print "hello world"
# I indented these lines
# 4 spaces, and now they show
# up as a code block.

print "hello world"
MathAppears as
Remember to wrap math in $$...$$ or $...$ to ensure proper formatting.
2 \times 3 $$2 \times 3$$
2^{34} $$2^{34}$$
a_{i-1} $$a_{i-1}$$
\frac{2}{3} $$\frac{2}{3}$$
\sqrt{2} $$\sqrt{2}$$
\sum_{i=1}^3 $$\sum_{i=1}^3$$
\sin \theta $$\sin \theta$$
\boxed{123} $$\boxed{123}$$

Sort by:

Use VECTORS.

- 4 years, 9 months ago

Torque is r X F. Take the cross-product.

- 4 years, 9 months ago

if it is parallel,and u happen to be strong enough as to break the axis of rotation,linear motion takes place

- 4 years, 9 months ago

If the force is not perpendicular, there are two ways of calculating the torque. I refer to this picture.

1. Splitting the force ($$F$$) into two components, one perpendicular to the axis ($$F_h$$) and one directly towards the axis ($$F_v$$). Then the torque $$\tau$$ becomes $$\tau=F_h l=Fl\sin(\theta)$$.

2. Projecting the arm so it's perpendicular to the force. Here, the torque becomes $$\tau=Fl'=Fl\sin(\theta)$$. As you see, the two methods agree, and are actually the same thing.

In more advanced physics, you usually calculate torque using vectors instead: $$\vec \tau=\vec l\times \vec F$$, where $$\|\vec \tau\|=\|\vec l\| \| \vec F\| \sin(\theta)$$.

- 4 years, 9 months ago

×