The diagram above shows a block of mass 1.3 kg, on a frictionless horizontal sur
ID: 2216323 • Letter: T
Question
The diagram above shows a block of mass 1.3 kg, on a frictionless horizontal surface, as seen from above. Three forces of magnitudes f2 = 7.03 N, f2= 4.04 N, and f3 = 8.99 N, are applied to the block, initially at rest on the surface, at angles shown on the diagram. In this problem, you will determine the resultant (total) force vector form the combination of the three individual force vectors. All angles should be measured counterclockwise from the positive x-axis (i.e. all angles are positive).
(1) Calculate the magnitude of the total resultant force acting on the mass.
(2) What anlge does Ftotal make with the positive x-axis?
(3) What is the magnitude of the mass's acceleration vector, a?
(4) What is the direction of a? In other words, what angle does this vector make with respect to the positive x-axis?
(5) How far (in meters) will the mass move in 5 seconds?
(6) What is the magnitude of the veloctiy vecotr of the block at t= 5 seconds?
(7) In what direction is the mass moving at time t = 5 seconds? That is what angle does the velocity vector make with respect to the positive x-axis?
Explanation / Answer
The Fr vector has an x and y component you can compute individually. The x-component is: (all angles relative to positive x-axis) Frx = F1*cos(25) + F2*cos(325) + F3*cos(180) (no calculator, sorry) y-component: (still using angles rel. to pos. x-axis, but now everything is in terms of sin function) Fry = F1*sin(25) + F2*sin(325) + F3*sin(180) So your Fr vector is a vector with those two components. When you compute it, you should see that it has a small downward component (negative y-component) but without computing it, I'm not sure by eyeballing if it's pos. or negative x-component. a) the angle will be arctan or inverse tan whichever is on your calculator arctan(Fry/Frx). It'll be something in between 270 and 360 degrees from the pos. x-axis. if the x-comp. was a pos. number, and in between 180 and 270 if x-comp. was a negative number b) Your Fr vector direction determines your acceleration direction, so the answer is exactly the same as (a). c) First compute the magnitude of the acceleration, but remember that force (newtons) = mass*acceleration: Magnitude of force: MagF = sqrt(Frx^2 + Fry^2) But force = mass*acceleration = 2*acceleration So MagAccel = MagF/2 In your problem there is no initial velocity, so the distance traveled is given by this formula: d = 1/2*a*t^2. so your answer is d = 1/2*MagAccel*5^2 = 1/2*MagAccel*25 d) The velocity vector again will still be the same as the acceleration. If the acceleration changed direction at all during the 5 seconds, then the velocity vector would change. But it doesn't, so it's exactly the same answer as (a) and (b).