The diagram below shows a block of mass m = 2.00kg on a frictionless horizontal
ID: 2061042 • Letter: T
Question
The diagram below shows a block of mass m = 2.00kg on a frictionless horizontal surface, as seen iron above Three forces of magnitudes F1=4.00 N F2=6.00 N and F3=8.00 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). What is the direction of ? In other words, what angle does this vector make with respect to the positive x axis? How far (In meters) will the mass move in 5.0 s? What is the magnitude of the velocity vector of the block at t =5.0 s ? In what direction is the mass mewing at time t=5.0s? That is, what angle does the velocity vector make with respect to the positive x axis?Explanation / Answer
the problem can be easily solved by using vectors.......
F1 = 4cos(25) i + 4 sin(25) j
F2 = 6cos(325) i + 6sin(325) j
F3 = 8cos(180) i + 8sin(180) j
resultant force F = F1+ F2+ F3
= 4cos(25) i + 4 sin(25) j + 6cos(325) i + 6sin(325) j + 8cos(180) i + 8sin(180) j
= (0.540 i - 1.75 j ) N
magnitude of resultant force = (0.5402 + 1.752)0.5 = 1.831 N
angle made by resultant force with x axis = tan-1(-1.75/0.540) = -72.851
magnitude of mass's acceleration 'a' = magnitude of resultant force/ mass
= 1.831/2 = 0.9155 m/s2
the direction of accleration will be the same as that of the resultant force i.e = -72.851