It was a little ambiguous.. But because the force acting on the handles has a vertical and horizontal component to it hes just solving for the perpendicular distance for each of those forces.. Hence "up then over". Hope that helps
@@aksccrdude But the force acting on the handles doesn't have a horizontal component does it? It's going in the negative y direction. So I understand the -100*cos(45)*(400) term, but I don't see where the +100*sin(15)*20 term comes from.
@@ignatiusjacquesreilly70 They are turning the 100 N into two components, one perpendicular and one parallel to the handle. When you do that, the force that is parallel to the handle has a distance of 20mm to F in the diagram and is contributing to the moment, whereas the other has a distance of 400mm. At Ay, the force that is along the handle is the only force that contributes a moment as the perpendicular force is inline/has zero distance from F
+Ajayi Nifemi newtons law, the sum of the forces in the x direction = 0 (if the machine is in static equilibrium aka not moving, (the title says this is a statics problem so its not moving)) so all the x direction forces added together = 0, since ax is only x direction force it is 0
That's beautiful. Thanks for the knowledge.
I don't understand how the horizontal distance from force a to force f is 20mm.
At around 5:20, the aprt when you said you went "up and then over" is kind of ambiguous to me.
It was a little ambiguous.. But because the force acting on the handles has a vertical and horizontal component to it hes just solving for the perpendicular distance for each of those forces.. Hence "up then over". Hope that helps
@@aksccrdude But the force acting on the handles doesn't have a horizontal component does it? It's going in the negative y direction. So I understand the -100*cos(45)*(400) term, but I don't see where the +100*sin(15)*20 term comes from.
@@ignatiusjacquesreilly70 They are turning the 100 N into two components, one perpendicular and one parallel to the handle. When you do that, the force that is parallel to the handle has a distance of 20mm to F in the diagram and is contributing to the moment, whereas the other has a distance of 400mm. At Ay, the force that is along the handle is the only force that contributes a moment as the perpendicular force is inline/has zero distance from F
I thought ay should be cos 15
I thought you couldnt sum the moments in ED ? Can someone clarify this for me?
When I sum the moments at Mf, I'm getting a different number than the 7000.
I'm gettiing 493 newtons
I am here after quiz lol
how is Ax 0?
+Ajayi Nifemi the total Fx is zero so, Ax is zero , because there is no force except Ax at the x direction.
+Ajayi Nifemi newtons law, the sum of the forces in the x direction = 0 (if the machine is in static equilibrium aka not moving, (the title says this is a statics problem so its not moving)) so all the x direction forces added together = 0, since ax is only x direction force it is 0
I don’t get the same Ay