Three forces of magnitudes F 1 = 4.0 N , F 2 = 6.0 N , and F 3 = 8.0 N are appli
ID: 2133074 • Letter: T
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).
A)
Explanation / Answer
Answer question
Fnetx = 4*cos25+6*cos325+8*cos180= 0.540 N
Fnet y = 4*sin25+6*sin325+8*sin180 = ?1.751 N
A) magnitude = 1.83 N
B) angle = 287.14
C) a = Fnet /m =0.915m/s^2
D )angle = 287.14
E) S = 0.5*a*t^2 = 11.44m