Recent content by Nemean

then it would be 30

it would be 15N

Normal force = mg - Force applied y (20*10) - 100sin(30) = 150N Fg = mg Fg = 200N

so it would be .2(100cos(30))?

.2(200cos(30)?

force applied = 100N

so wouldn't that be Fx = Facos(30) - T

O right, the applied force. Now would i add that to what i had gotten before? so 100cos(30) ?

well that's were i get a bit lost. I need to find the net force, and i figured that if i used the x component of box 1 and box 2 so i got 200cos30 for box one, and 400sin30 for box two

If i use (.2)(200cos(30)) + (.2)(400sin(30)) only equals 74N and the answer is 63.8

i believe they are moving to the left

Homework Statement Find the Net force on the system see thumbnail for given. Normal force action on object 1 - 150N Normal force action on object 2 - 346N Force of friction action on object #1 - 15N Force of friction action on object #2 - 34.6N M1 moves to the left Homework Equations F = ma...