QI/ Find the magnitude and direction of the resultant forces as shown in figure 1, when e-32° ? 200N 60° Figurel 100 N
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- Resolve the 360-lb force into components along the cables AB and AC. Use =60 and =40.Given r=4i6j+2km (position vector) F=20i+40j30kN (force vector) =0.8j+0.6k (dimensionless unit vector) compute (a) rF; and (b) rF.Determine (a) the angle between the position vectors P and Q; and (b) a unit vector perpendicular to P and Q.
- The force systems in Figs. (a) and (b) have the same moment about the x- and y-axis. Determine the distances x and y.The typical power output of a compact car engine is 110 hp. What is the equivalent power in (a) lbft/s; and (b) kW?The length of the position vector r is 240 mm. Determine the rectangular components of (a) r; and (b) the unit vector directed from O toward A.
- It can be show that a plane area may he represented by a vector A=A, where A is the area and represents a unit vector normal to the piano: of the area. Show that the area vector of the parallelogram formed by the vectors a and b shown in the figure is A=ab.1.2) Problem. 3 The following forces (all pull) act at a point : (i) 25 N due North ; (ii) 10 N North-East ; (iii) 15 N due East ; (iv) 20 N 30° East of South ; (v) 30 N 60° South of West. Find the resultant force. What angle does it make with East ?Four parallel forces act on the plate as shown. For the force system shown, find the equivalent resultant force and specify its x and y location from the origin, given: FA = 170 lbs, FB = 240 lbs, FC = 210 lbs, FD = 140 lbs, LX = 16 ft, LY = 14 ft, BY = 5 ft, DX = 6 ft The magnitude of resultant force, FR is: (positive number) FR = lbs The distance from the Origin in the positive X-direction is: X = ft The distance from the Origin in the positive Y-direction is: Y = ft