Three forces A (30N at 35 degrees NE), B (40N at 25 degrees SW) and C (50N at SE) all act on one original point. Find the magnitude and direction measured from the positive x-axis of a fourth force F required to keep the body in equilibrium.
3 answers
What is the direction of the 3rd force?
Sorry . for the force A , the direction is actually NW and the direction for C is 40 degrees SW
All angles are measured CCW from the +x-axis.
Fr = 30N[145o] + 40[205o] + 50[220o].
Fr = (-24.57+17.21i) + (-36.25-16.90i) + (-38.30-32.14i)
-99.12 - -31.83i = 104.11N[17.8o] S. of W. = 104.11N[197.8o]CCW. = resultant force.
To keep the body in equilibrium, F4 must be equal in magnitude and 180o out of phase(17.8o N. of E. or 17.8o CCW) With Fr.
F4 = 104.11N[17.8o]CCW.
Fr = 30N[145o] + 40[205o] + 50[220o].
Fr = (-24.57+17.21i) + (-36.25-16.90i) + (-38.30-32.14i)
-99.12 - -31.83i = 104.11N[17.8o] S. of W. = 104.11N[197.8o]CCW. = resultant force.
To keep the body in equilibrium, F4 must be equal in magnitude and 180o out of phase(17.8o N. of E. or 17.8o CCW) With Fr.
F4 = 104.11N[17.8o]CCW.
1 answer