Question

I got part A correct with an answer of 7.96 kN. I cannot figure out part B

If F- 7.2 kN, determine the magnitude of the resultant force acting on the screw eye 45° 60° 2 kN Express your answer to three significant figures and include the appropriate units

Answer 1

The force vector diagram is as follows, 2 kN Part A F = 7.2 kN The angle between the two forces is, 0=180° -60° - 45° = 75° The magnitude of the resultant force acting on the screw eye is, R = F? +(2kN)* + 2F (2kN) cos e = V(7.2 kN)? +(2kN)? +2(7.2 kN) (2kN) cos 75° - 72 KN ( 1983) + ( 2kN ( 1998) +2( 72 kN( 1969) (22N( 1965N) (025851) = (1000 N) (72 N) +(2N)2 + 2(7.2 N)(2N)(0.25881) =(1000 N)(7.95573 N) = 7955.73 N 1 kN) 9. 1000 N = 7.95 kN Therefore, the magnitude of the resultant force acting on the screw eye is 7.95 kN).

tanas. Part B The angle between the two force from the force 2 kN is, Fsino (2 kN)+F cose (7.2 N sin 75° +(2kN)+(7.2 N)cos 75° (7.2 kN)(0.9659) (2 kN)+(7.2 kN)(0.25881) =1.800 a = tan(1.800) = 60.94° The direction of the resultant force acting on the screw eye measured clockwise from the x axis is, = a +45° = 60.94° +45° = 105.94° Therefore, the direction of the resultant force acting on the screw eye in clockwise from the x axis is 105.94°