Results 1 to 4 of 4

Math Help - Vectors and forces

  1. #1
    Newbie
    Joined
    Sep 2012
    From
    brisbane
    Posts
    2

    Vectors and forces

    Hi.
    This is the first time I've posted in this forum.
    I'm completely stumped on a question to do with applications of vectors and forces.
    If a 50kg seesaw is at rest on a plane inclined at 30 degrees to the horizontal, determine the forces acting on the seesaw.

    Greatly appreciate any help.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor
    Joined
    Sep 2012
    From
    Australia
    Posts
    4,185
    Thanks
    772

    Re: Vectors and forces

    Hey tomato89.

    This is going to be a little messy if you did a real world example due to the distribution of mass of the see-saw and how this affects the rest of it, but if you treat it has some standard object, you consider the net force of that object moving and how to decompose it.

    So you will have a gravitational force of the see-saw acting in the direction it acts (which is approximated to be negative y axis) and you decompose that force into the x and y components by doing a projection onto the <1,0> and <0,1> vectors. Same for friction (find out the friction magnitude in the direction of the incline opposing the object and decompose it into x and y parts).

    Now you've got the forces provided by the thing the see-saw is sitting on which is that plane. You've got the gravitational forces of the see-saw that will give it a net force down the plane. If the frictional and object forces don't completely cancel this out, then the object will accelerate.

    Now what can happen in these toy examples is that we only consider the net-force in the direction opposing the frictional force and disgregard the rest.

    So if you have a gravitational force acting in the negative y direction with some magnitude, you need to find the component acting in the direction that is parallel to the inclined plane as it heads down.

    When you add the two opposing forces together (they will be two vectors but they will point in the opposite directions) you will get a net force. If that force is non-zero, the object will accelerate.

    So if you have a gravitational force down, it has a magnitude of mg where g = 9.8 m/s^2. and m is 50. So now you have to find the component of that force in the direction going down the plane. This is a projection.

    You said the plane is inclined at 30 degrees to the horizontal which means that you have to project your vector that is <0,-50*9.8> onto the vector that is counterclockwise of 30+180 degrees from the x-axis. This vector is basically <cos(210),sin(210)> which has a length 1.

    If you find the length of the gravitational component in that direction, you get the magnitude to be 0*cos(210) + -50*9.8*sin(210) = -50*9.8*-sin(30) = 50*9.8*sin(30) which is usually a trick you use in these inclined problems.

    That's the real derivation of how you get this result: it is simply a projection operation of projecting one vector onto another.

    So now you need to get the magnitude of the friction but you make this negative since it is going in the opposite direction (but it will always be between 0 and the magnitude you just calculated above).

    If it is non-zero you now have to project this vector back onto the x and y vectors. The x vector is <1,0> and the y vector is <0,1>. The original vector though is <cos(210),sin(210)> but it has a magnitude of whatever you got (call it m) so your new forces in x and y co-ordinates will be m*cos(210) Newtons in the x direction and m*sin(210) in the y direction, and since cos(210) = -cos(30) and sin(210) = -sin(30) you have the x component being -m*sin(30) for x and -m*cos(30) for y which is intuitive since it means it is accelerating both left and downwards which is what you would expect for an inclined plane heading in that direction.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Super Member
    Joined
    Sep 2012
    From
    Washington DC USA
    Posts
    525
    Thanks
    147

    Re: Vectors and forces

    Whenever you set up a problem with something "resting on" something else, you have what's called a "normal force". What that is is a force to show that the thing doesn't move through the object on which it's resting.

    Ex: A man stands on the surface of the Earth. What are the forces acting on him? If you only say gravity, then the man has a net force acting on him, and so he hurls toward the center of the Earth! There are two forces acting on him - gravity, AND the normal force from the surface of the Earth pushing up on him, keeping him from falling toward the center of the Earth. The *net* force acting on him is zero - the normal force exactly balances the force of gravity. If the net force weren't zero, he'd be accelerating - but he of course doesn't accelerate.

    The normal force is always perpenticular to the "contraint", to the thing that, because it's solid, the probelm says you can't pass through it. It's magnitude is exactly enough to exactly balance component of the net of all other forces in that direction. It's essentially a catch-all "eraser" that erases any net component of force that would otherwise result in you falling through the surface.

    In this problem, the forces are:
    1) The force of gravity: direction is straight down, magnitude is mg.
    2) The normal force: direction is perpendicular to the plane, magnitude is whatever it needs to be so that the *net* force in the direction perpendicular to the plane is 0 (there's no acceleration blasting through the plane).
    3) Friction: direction is always in opposition to the direction of motion, magnitude depends, BUT for these "sliding on an inclined plane" problems, is a number (called the coefficient of friction - comes in resting and moving varieties) that when multiplied by the magnitude of the normal force gives you the magnitude of the frictional force. (For the resting case, so long as the magnitude of the component of gravity parallel to the incline is less than this magnitude computed for friction, there's no "beginning" to move, so the actual magnitude of the force of friction will be exactly that necessary to balance the component of the force of gravity in order that no motion takes place - i.e. in order that the net force on the still-standing object is 0.)

    Then you break the force of gravity into it's components in parallel and perpendicular to the plane. The normal force counters the perpendicular component. That magnitude allows you to compute the force of friction. End up with a net force parallel to the object, gravity pushing it down the incline, and friction opposing it.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Newbie
    Joined
    Sep 2012
    From
    brisbane
    Posts
    2

    Re: Vectors and forces

    Thank you so much. I've done the problems that have something resting on the seesaw, but I couldn't get my head around the forces I have to work out when the seesaw is isolated. Thanks again
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Vectors, Forces and Moments
    Posted in the Algebra Forum
    Replies: 0
    Last Post: November 26th 2011, 09:48 AM
  2. Vectors, Forces and Moments
    Posted in the Math Topics Forum
    Replies: 0
    Last Post: November 26th 2011, 09:48 AM
  3. Forces in Equilibrium (Vectors) II
    Posted in the Advanced Applied Math Forum
    Replies: 14
    Last Post: October 2nd 2008, 06:43 PM
  4. Forces and Vectors
    Posted in the Trigonometry Forum
    Replies: 4
    Last Post: August 18th 2008, 04:43 AM
  5. Forces as a Vectors
    Posted in the Advanced Applied Math Forum
    Replies: 2
    Last Post: February 14th 2008, 04:17 PM

Search Tags


/mathhelpforum @mathhelpforum