Recent content by Moayad

there is four way to antiderivative (as i have known ) ... first one is subtution second using the table of intgration third is by part fourth is friction excuse my splling

case one i-Fnet,x= FA cos(0)+FBcos(0)=ma 0- because they are in the orginal x-axis and pointed to the right a=0.57 m/s2 case two do the same thing what i did in case one Fnet,x= FA cos(0)+FA cos(180)=ma,,, (a here is .35) 180 because it is ponited to the left u have two...

i got it ... thank you for you help i used the three equation 1- F=sqrt((Fx)^2+(Fy)^2) F=120 2-Fx = -100 cos(30)-30 cos(θ)-30cos(θ+50) 3-Fy=100sin(30)-30sin(θ)-30sin(θ+50) i pluged equation 2,3 to 1 and solve for θ i got 27.39>θ>223

i used 30/sin(25)=R/sin(40) R= the new force

Hi again ... i usually make stupid mistakes ... sorry heh i added them so what i got is 45.69 lb i am going to work this out .. and i'll be back thank you sooooo much

here again i did this : i used the x to the right, so Fx=-100cos(30)-30cos(25-θ) , ( i combined the two 30lbs forces into one as you told me ) Fy=100sin(30)-30sin(25+θ) F^2 = Fx^2 + Fy^2 l let F=120 ...(is it right?) so, Fx= sqrt((120)^2+Fy^2) i pulged to...

from the graph is not the x to the right ? sorry i did not use what you gave me, i' am going to use from now thank you again

thanks man ... can i ask more questions is not Fnet,x = -100 cos(30)+30 cos(@)+30cos(@+50)=0 what about Fnet,y? is it goin to be = 100sin(30)+30sin(@)+30sin(@+50)? Thank you for help Moayad was here

Homework Statement Three forces are applied to a bracket. The directions of the two 30-lb forces can vary, but the angle between these forces is always 50. Determine the range of values of  for which the magnitude of the resultant of the forces applied to the bracket is less than 120...