Question

Please do not answer if you do not know how to get the correct answer! I've posted 15 times with no correct answer! a) is 164.4 N. Thanks!

A symmetrical table of height 0.750 m, length 1.55 m, and weight 462 N is dragged across the floor by a force applied to its front edge. The force is directed to the right and upward and makes an angle of 30.0° with the horizontal.

(a) Find the minimum force necessary to drag the table across the floor. The coefficient of sliding friction between table and floor is 0.375.
  

(b) Calculate the normal force and the frictional force on each leg.

front legs:	normal force
	frictional force
back legs:	normal force
	frictional force

Answer 1

oluti on F(bう 0150 630 2 eltie Ne trge on the tableO &.kon Consid 九an ani.

(%)(1.55)-(Fade) (0.7 50 )-(462)(L55) =0 2. ny (and 46 2 N2= 162.1 N - 81,6 y N ente mnl force on ea = 4-0, 82 N nal force on each-toan le gs 30, yN