Chapter 4, Problem 95P

Q6- Two forces F1=240 lb and F2=160lb act on the hip joint. Determine the resultant of the two forces and express the result as a cartesian vector. Also, find the coordinate direction angles of the resultant force. Z F1= 240 lb 75° 131 F2 = 160 lb ww 45° y

Q: Represent the resultant of the three forces and one couple by an 10 points equivalent force R at A and a couple M. find M and the magnitude and direction of R* +30 (400 lb 1000 lb-ft 800 300 lb

3- If the two forces F, (100i - 120j + 75k) Ib and F, = {-200i + 250j + 100k) lb act at A, determine the resultant moment produced by these forces about point O. Express the result as a Cartesian vector.

3. Replace the two forces acting on the post by a resultant force and couple moment at point O. Express the results in Cartesian vector form. C FB=5 kN Fp =7 kN 6 m 8m 2 m 3 m. 6 m.

2/35 The 30-N force P is applied perpendicular to the por- tion BC of the bent bar. Determine the moments of P about point B and about point A.

If the couple moment has a magnitude of 300 lb-ft, determine the magnitude of F of the couple forces 12 ft -F 7 ft 2 ft 1 ft 6ft F -12 ft-

Calculate the resultant moment produced by two forces about point O. Express the result as a Cartesian vector.

Determine the coordinate direction angles of the force F applied at the end of the pipe such that the moment of F about point A is zero.

Determine the moment of the 80-lb force about the Oa axis. Express the result as a Cartesian vector. C 5 ft 30° 4 ft B 30° 45° 80 lb 6 ft 60° 120°