The diagram below shows a block of mass m=2.00\\; \ m kg on a frictionless horiz
ID: 2042612 • Letter: T
Question
The diagram below shows a block of mass m=2.00; m kg on a frictionless horizontal surface, as seen from above. Three forces of magnitudes F_1 = 4.00;{ m N}, F_2 = 6.00;{ m N}, and F_3 = 8.00;{ m 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 from 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).Explanation / Answer
1)Fx = F1x + F2x +F3x = (4N)(cos25o)+ (6N)(cos325o)+ (8N)(cos180o) = 3.625N + 4.915N - 8N =0.54N Fy = F1y +F2y + F3y =(4N)(sin25o)+ (6N)(sin325o) +(8N)(sin180o) = 1.69N -3.44N + 0N =-1.75N Then teh magnitude of resultant force is Fr =sqrt[(0.54N)2 + (-1.75N)2 ] = 1.83N 2) Direction is ? =tan-1[-1.75/0.54] = 180o+ tan-1[1.75/0.54] = 252.85o 3) acceleration a = (1.83N)/(2kg) = 0.915m/s2 4) The direction of acceleration is the direction of force= 252.85o 5) v = (0.915m/s2)(5s) = 4.575m/s 6) The direction of velocity is the direction of force= 252.85o