The normal force must act on the box, and as x is measured from the center of the box, more than half of the width will put the normal force, N off the box.
@@engineeringvision Respectfully, I don't believe this is correct. Your calculation on the sum of moments changed units from feet to meters. If you're determining the smallest possible force to get the box to move, there are two possible answers, sliding movement or tipping movement. When calculating the tipping, take a sum of moments about the bottom left corner, which is what the box would rotate about, rather than in the bottom center. At the exact moment of tip, the normal force would be at the only point of contact between the box and the surface- the bottom left corner- and therefore eliminated from the equation. M=P(3')-150(1')=0. Lowest possible force to move the box is 50lb.
why it is not possible for x to be greater than half the width of the crate?
The normal force must act on the box, and as x is measured from the center of the box, more than half of the width will put the normal force, N off the box.
@@engineeringvision Respectfully, I don't believe this is correct. Your calculation on the sum of moments changed units from feet to meters. If you're determining the smallest possible force to get the box to move, there are two possible answers, sliding movement or tipping movement. When calculating the tipping, take a sum of moments about the bottom left corner, which is what the box would rotate about, rather than in the bottom center. At the exact moment of tip, the normal force would be at the only point of contact between the box and the surface- the bottom left corner- and therefore eliminated from the equation. M=P(3')-150(1')=0. Lowest possible force to move the box is 50lb.
@@WeDoDumbStuffit's correct. Just he said meters instead of feet