Understanding Forces and their Vector Components

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I was told forces are vectors so they can be divided into x and y components, but what about the normal force or the force of static friction? do you divide those into x and y components if say you had a block that was lying on an angled surface?
Relevant Equations
Fn = mg
I was told forces are vectors so they can be divided into x and y components, but what about the normal force or the force of static friction? do you divide those into x and y components if say you had a block that was lying on an angled surface?
 
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Yes, any vector can be resolved into x and y (and z) components. Sometimes one or more of those components is zero - for example if you have a vector parallel to the x axis its components in y- and z-directions will be zero.
 
random124 said:
I was told forces are vectors so they can be divided into x and y components, but what about the normal force or the force of static friction? do you divide those into x and y components if say you had a block that was lying on an angled surface?
In the case of a block lying on the surface at rest, the normal force and the force of static friction are components of the contact force exerted by the surface. The component of this contact force that is parallel to the surface is also known as the force of static friction and the component perpendicular to the surface is also known as the normal force. The vector sum of these two force has magnitude equal to the weight of the block and direction straight up.
 
I like to say "every vector is the hypotenuse of some right triangle, with legs parallel to a pair of perpendicular axes".
But, Physics doesn't care which pair you use.
However, for some pairs, it may lead to a simpler math problem (e.g. simpler system of equations), or something easier to interpret physically (e.g. normal and friction for surfaces in contact).

For linear motion on an inclined surface, one often chooses axes parallel and perpendicuar to the surface.

For circular motion on a banked surface, one often chooses axes parallel and perpendicular to the axis of rotation.
 
random124 said:
but what about the normal force or the force of static friction?
do you divide those into x and y components if say you had a block that was lying on an angled surface?
First, choose your axes. You don't have to have x as horizontal and y as vertical. Sometimes it makes the algebra easier if you choose, say, x parallel to a slope and y normal to it. (Making them not orthogonal to each other makes things messy.)

The "normal force" between two bodies is actually a pair of equal and opposite forces the bodies exert on each other. They are the minimum magnitude forces that will prevent interpenetration. It follows from that that they act normally to the contact plane, hence the name.
 
random124 said:
or the force of static friction? do you divide those into x and y components if say you had a block that was lying on an angled surface?
This is an interesting question. Think about a car. There is a vertical component of the contact force that supports the car's weight. But, how does a car turn? Turning requires a force. And, that force is provided by static friction between the tyres and the road. The contact force, therefore, has all three components. The z component is the car's weight. And the x and y components depend on the car's change of direction. Or, indeed, on any form of acceleration.
 
Thread 'Variable mass system : water sprayed into a moving container'
Starting with the mass considerations #m(t)# is mass of water #M_{c}# mass of container and #M(t)# mass of total system $$M(t) = M_{C} + m(t)$$ $$\Rightarrow \frac{dM(t)}{dt} = \frac{dm(t)}{dt}$$ $$P_i = Mv + u \, dm$$ $$P_f = (M + dm)(v + dv)$$ $$\Delta P = M \, dv + (v - u) \, dm$$ $$F = \frac{dP}{dt} = M \frac{dv}{dt} + (v - u) \frac{dm}{dt}$$ $$F = u \frac{dm}{dt} = \rho A u^2$$ from conservation of momentum , the cannon recoils with the same force which it applies. $$\quad \frac{dm}{dt}...
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