Equilibrium in non co-planar system

I got this question while I was solving questions from an entrance book.... Q. Find the minimum number of forces required to keep a body in equilibrium in a non co-planar system. Can anyone help me in this problem??

#Physics

Easy Math Editor

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Comments

Presumably the "no force" answer is not acceptable, since a system with no force acting is a planar system.

You have to have at least three forces to have a non-planar problem, and the sum of three nonplanar forces cannot be zero, and hence the body will not be in equilibrium. Thus you need at least four forces for equilibrium.

If you consider an object being acted on by four forces, equal in magnitude, that are pointed at the object from the four vertices of a regular tetrahedron, with the object at the centre, you see that we can obtain equilibrium with four forces. What you need is four forces that add to zero, and whose lines of action are collinear (so that their moment about the point where the lines meet is zero).

Mark Hennings - 7 years, 6 months ago

Now I understood the concept. Thanks a lot sir.

A Former Brilliant Member - 7 years, 6 months ago

Zero. Do not apply any force at all. If you would like to apply one, then you need another exactly opposite and equal in magnitude.

SAMARTH M.O. - 7 years, 7 months ago

Force =0 since. Body still at rest

lawal Adeshina - 7 years, 7 months ago

Math	Appears 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}`	$\boxed{123}$