octrola markys jaylie
Maureen Quartaro
Problem &: The diagram shows the two forces with magnitudes F and Fz acting on a body. The first force is in the positive direction, while the second makes an angle with the negative direction as shown.
2) For part 2, use Phet simulation for each vector to display the vector components and then fill in Fx, Fy, Fzx, Fzy, Fss, Fsy, FExs, Fo, IFr, and the angle where the equilibrant force makes with the positive X-axis in table 2. Calculate the percentage error of the magnitude of the experimental value of FR compared to the analytical solution for FR.
For part use Phet simulation for each vector to display the vector components and then fill in the values for Fx, Fy, FEx, FEy, Fgl, and the angle where the equilibrant force makes with the positive X-axis in the table. Calculate the percentage error of the magnitude of the experimental value of Fr compared to the analytical solution for Fr.
Vetrossan-c
2- Resultant of three vectors: With the first vector F fixed at approximately 30 degrees and its magnitude is 6 N. The second vector F fixed at an angle of approximately 145 degrees and its value is 11 N. Find the resultant force of these three forces using Phet simulation F FR.
OR
Write down the value of Fs and its direction.
FEF
PHET
OF
0
Data Analysis:
PHET: ta
1. For part 1, use Phet simulation for each vector to display the vector's components and then fill Fx, Fy, and the angle where the equilibrant force makes with the positive x-axis in table 1. a. Calculate the percentage error of the magnitude of the experimental value of Fg compared to the analytical solution for F.
Calculate the resultant force of adding " the two forces F Fz using the data from phet simulation, then write your results in table (2). Table (2): analytical method Force Magnitude (N) Direction (degree) component Y-component
What is the relation between the resultant force FR (FR = F Fz) and the balancing force FB?
Calculate the percentage difference for the resultant force FR between the results from analytical (step 4) and graphical methods (step 3).
So, according to Arthur C. Clarke, famous author and originator of the idea of the geosynchronous communications satellite, perpetual motion is no big deal. What is a big deal, of course, is arranging for such perpetual motion to occur on Earth. This is a near impossibilty, although we can get close: Clarke suggests magnetically suspending a heavy flywheel in a vacuum. (He also goes on to make the point that perpetual motion machines that can do useful external work while they run are, in fact, utterly impossible.)
However, the point of the First Law is to act as the foundation stone of Newtonian dynamics: any deviation of an object from a straight line path is taken as implying the existence of a resultant force; if there is no deviation there no force and vice versa.
What we see is the ball oscillating back and forth along the track. However, what we also observe is that the height reached by the ball gradually decreases. This is because of resistive forces that slow down the ball (e.g. friction between the track and the ball and air resistance).
Now that i think about it, the answer is 1. 1 could not have an obtainable equilibrium. When you put the other ones at funky angles you can always achieve an equilibrium. BUT, for number 1, even if you had the 5N force going one way, and the 1N and 3N force ON TOP of eachother heading in the exact opposite direction, you could not add up to the 5N needed to attain an equilibrium. Adding an angle between the forces only reduces the resultant (there is no sinx or cosx value bigger then 1).
So to do this problem, the first step would be to see if 2 of the resultants can add up to be greater then or equal to the other resultant in all cases. For number 1, 1N + 3N < 5N, so it would be impossible to have an equilibrium
You might have to play for a bit to determine the angles (the math gets a bit more involved), but it's pretty straightforward to see how you could line up a 4N, 4N, and a 5N vector to achieve equilibrium... I've used the vector simulation lab from PHET ( ) to demonstrate how putting a 4N, 4N, and 5N vector together gives you a vector sum of 0.
Okay this time I am in dire need of help - A 50N horizontal force is needed to keep an object weighing 500 N moving at a constant velocity of 2 m/s across a horizontal surface. what is the magnitude of the frictional force acting on the object?
If the object is at constant velocity (that means it is not accelerating) then Fnet= 0 since F=ma and a=0 at constant velocity. If it takes 50N to pull it at constant velocity, then the force of friction must be 50N in the opposite direction for Fnet to be 0!
not if the object accelerated until it got to 2 m/s but when it reached that point the force would have had to change to a point where it equaled the force of friction. It doesnt say anything about what happend before 2 m/s so you dont have to worry about that
Remember, no net force doesn't mean an object is at rest. no net force means an object won't accelerate -- therefore an object at rest will remain at rest, and an object in motion will remain in motion at a constant velocity and in a straight line -- unless there is a net force.
For example, if a spaceship is moving at 1000 m/s through space, if Superman pushes it forward with a force of 50N and The Hulk pushes it backward with a force of 50N, the forces are balanced, so there is no net force, therefore the spaceship continues at 1000 m/s. Net force = 0 implies no acceleration.
If you weren't a super duper physics student, it would seem so! However, Forces cause accelerations on an object. and if an object is not accelerating in any direction, two things can be happening to the object. It can either be motionless OR in constant motion; talk to yourself about the definition of constant velocity, thats how I wrapped my brain around that one. Velocity only changes with acceleration (forces), so a constant velocity means no forces or forces that cancel each other out. I hope this helps!
It would seem so. What you have to keep in mind Newton's Second Law of Physics F=ma. A force results only in an acceleration. AND we know that if an object has Fnet=0 (that means that there is either no forces acting on an object, or that the forces acting on the object cancel each other, as in this case) then the object is not accelerating in any direction. There are only two possibilities for an object that is experiencing no accelerations; this comes from Newton's First Law: an object will remain at rest or in constant motion unless acted upon by an outside force. Your object, however, also has a velocity. Ergo, your object is moving with constant velocity. Hope this helps!
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