Question

Three forces of magnitudes F1=4.0N, F2=6.0N, and F3=8.0N are applied to a block of mass m=2.0kg, initially at rest, at angles shown on the diagram. (Figure 1) In this problem, you will determine the resultant (net) force by combining the three individual force vectors. All angles should be measured counterclockwise from the positive x axis (i.e., all angles are positive).

calculate the magnitude of the resultant forceF?r=F?1+F?2+F?3 acting on the block.

What angle does F?r make with the positive xaxis?

What is the magnitude of the block's acceleration vector,a??

What is the direction of a?? In other words, what angle does this vector make with respect to the positive xaxis?

How far (in meters) will the block move in 5.0 s? Recall that it starts from rest.

Answer 1

you should be breaking each force into their horizontal and vertical components. Thats what you get when you multipy the force by cosine of the ange (horizontal component) or by the sine of the ange (vertical component). Because the components of these forces are not all in the same direction, the sign (+-) is important.

But you've done all that, and you find that the resultant force has a horizontal component, Fx = .5401 N and a vertical component, Fy = 1.75099 N. (Fy points up, and Fx points left (or in the opposite direction of F3))
The best way to think about the force is as a right triangle, with three sides, the hypotenuse being the net force, Fr and the vertical and horizontal sides being the vertical and horizontal components.
If you make the lengths of the sides equal to the magnitudes of the horizontal and vertical components, Fx and Fy, then the hypotenuse will be eqaual to the magnitude of the net force, Fr.
Likewise, if the triange is oriented correctly (having vertical and horizontal sides makes that easy) the hypotenuse will be pointing in the same direction as the net force.

Now that you have your triangle drawn, the next two steps should be easy. Just use the pythagorean theorem to find the length of the hypotenuse (which is equal to the magnitude of Fr) and if you know anything about trig functions, you know that tan(angle) = opposite side / adjacent side. In the case of this triangle, that means that tan (angle) = Fy / Fx.
Thus angle = inverse tan (Fy/Fx). (inverse tan is TAN^-1)
You should find, here that the angle between the horizontal side of the triange (and the x-axis) = -72.85 (it is below the x axis so it would also be the equivalent of 360 - 72.85 = 287.14)

The next problem asks for the magnitude of the block's acceleration vector, a. This is pretty easy. Newton's law states that F = m * a (mass times acceleration).
You know the force on the block (you're using magnitudes, so it don't need the angle), Fr=1.83 N. You also are given the mass of the block, so you can plug in both and find 1.83N = 2kg * a
divide both sides by 2kg, and you find that 1.83 N/ 2kg = your answer, .915 m/(s^2).
The next problem asks for the angle between the acceleration vector and the x-axis. This is easy, there is only one force acting on the block, so the acceleration will be in the same direction as the foce, thus -72.85 degrees or 287.14 degrees.

Next you move on to movement problems, the primary equation for which is X = 1/2 a* t^2 + (Vo)*t + Xo,
where X is the position, a is the acceleration, t is the time, Vo is the initial velocity, and Xo is the initial position.
With a Vo=0 and setting Xo=0 (because you're really trying to find X-Xo or the distance traveled, and thus X = X - Xo if Xo=0)
You equation is now X=1/2 a*t^2 .
plug in for the acceleration we found and for the time (5.0 s) and you find the distance traveled in 5 s.

Now that I look at problem 6, I'm not entirely sure what is being asked. Based on the angles given (F2: 325 degrees?) the net force would have been pointing up and two the left, thus the block would be being picked up. I'm taking this to mean that we're ignoring gravity for this equation (otherwise the gravity will overwhelm the vertical component of the force, as the force of gravity would be = 19.6 N.)

Other possibilities are that the way I'm interpreting the angles is opposite the way you, and hence your teacher, are. In this case the net force, Fr would be pointing into the ground. This would mean that there would be one additional force, one called the normal force, which is basically the force the ground applies up to prevent the block from going through it. The net force points straight up and negates the downwards vertical component of the net force. Thus the only force reamining on the block would be the horizontal component of the force Fr, which is Fx, which we found a long time ago. In this case, you would use the same movement formula, but replace Fr with Fx.

In either case the answers to 6 and 7 will be found the same way.

Assuming no other forces act on the block between t=0 and t=5, the velocity will be in the same direction as the acceleration and thus in the same direction as the force. The formula for velocity is v=a*t + Vo (Vo is the initial velocity, which is 0)
Thus you shouldn't find difficulty in plugging in whichever acceleration is correct, and t=5s.
The direction of the velocity at 5s is the same as the acceleration has been in. In either case, there has only been one direction for net forces, and thus only one direction for accelerations, which means that the direction of the velocity will be the same as the original acceleration (either 0 or -72.856)