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Inclined plane showing two vector diagrams

I've seen a number of examples which show an object on a friction-less plane. A vector diagram is used to show that a component of force due to gravity, -Fr, is balanced by the the reaction force (normal force) at 90 deg to the plane, Fr. This leaves Fx, the resultant force parallel to the plane. - see the left hand diagram at the bottom.

What if the size of Fr was drawn in the diagram larger than the size of Fg (see the right hand diagram)? The new resultant force, Fx, is now perpendicular to Fg. It has to be drawn in a vector diagram in the 'shape' of a parallelogram.

Can non-rectangular vector diagrams be used to find the resultant force on the object, Fx?

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Can non-rectangular vector diagrams be used to find the resultant force on the object, Fx?

Yes, you can decompose any vector in any basis you like, including a non-orthogonal basis. See the example below:

Non-orthonormal basis

But in most cases, including your example, that's far from practical.

In your example you want to determine the acceleration the object will experience. We know that assuming the incline is strong enough to resist the weight of the object, then the acceleration vector will be parallel to the incline.

In your LHS vector diagram you defined the $x$-axis in precisely that direction.

Newton's second then tells us:

$$ma_x=F_x$$

But in your RHS vector diagram you defined the $x$-axis as the horizontal. That's not the direction of acceleration, so you will still need to find the force component in that direction.

For problems like yours, choose the easiest to use coordinate system, which is almost always orthogonal (Cartesian).

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  • $\begingroup$ Thanks for the answer. I understand that the LH diagram is simpler but just for proof of concept - would the -Fr vector still cancel Fr in the RHS diagram? $\endgroup$ – Adrian Nov 15 '16 at 21:45
  • $\begingroup$ Yes, the diagram is correct: but not very useful! $\endgroup$ – Gert Nov 15 '16 at 21:50
  • $\begingroup$ @Adrian No, it wouldn't. If it did, the only leftover force would be the horizontal Fx force, which implies that the block would accelerate horizontally. But it must accelerate parallel to the plane. $\endgroup$ – knzhou Nov 15 '16 at 21:50
  • $\begingroup$ The reason is that there's still a component of Fx that points along the normal direction. So you would have to resolve Fx into components again to find the normal force. $\endgroup$ – knzhou Nov 15 '16 at 21:51
  • $\begingroup$ @Gert, but you would need the component parallel to the plane, right? $\endgroup$ – Adrian Nov 15 '16 at 21:54
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Yes, you can always decompose your forces any way you want.

However, in almost all intro mechanics problems, the best way is to decompose them into orthogonal components (what you call a 'rectangular vector'). The reason is that many of these problems take place in two dimensions with a constraint -- for example, your problem is two-dimensional, and the constraint is 'the block can't sink into the plane', which is enforced by the normal force.

Because of this constraint, the block can only move perpendicular to the plane. Therefore it's most practical to resolve the forces into components pointing straight into the plane, and perpendicular to it. The former determines the normal force, and the latter determines the block's acceleration.

Similarly, when you have a mass on a string, like on a pendulum, the constraint is often 'the string doesn't stretch'. Again, that means the motion is perpendicular to the string, and you want to resolve forces into components parallel to the string (like tension) and perpendicular to it.

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