Not considering friction or other forces, how much force, pressure, would it take to push, lift, a 1200 lb. weight up a 45 deg. angle.
I am curious as I saw a weight lifter do this using his legs and thought he was not actually lifting a 1200 lb. resistance.
2006-11-27 05:30:59 · 4 answers · asked by tronary 7 in Science & Mathematics ➔ Engineering
The component of the force acting down the ramp would be W x sin (45) or .707 x 1200 = about 850 lb.
But you can't neglect friction a reasonable number would be 20% of the component of the force acting at 90 degrees to the ramp or W x cos 45 = (.20) x (1200) x 0.707 = about 170 lb
For a total load of 1020 lbs.
2006-11-27 05:39:19 · answer #1 · answered by Roadkill 6 · 0⤊ 0⤋
Neglecting friction, he is lifting an equivalent of sin(45 degrees) times the weight. The math is as follows:
The weight the lifter sees = sine of angle between the slope and level ground * weight of mass
In this case, sine of 45 degrees is 0.707.
Thanks for pointing out my mistake. Sin of 45 is .707 not 0.5.
2006-11-27 13:40:23 · answer #2 · answered by Anonymous · 0⤊ 0⤋
The sine of 45° is actually .707. He is lifting 70% of the weight.
Never trust a snow-boarder with trig....
2006-11-27 22:27:11 · answer #3 · answered by www.HaysEngineering.com 4 · 0⤊ 0⤋
Hello.
Riiiight.. Well I suck at math.
Soo, good luck with that one!
I can lift 1200:D
Hah.. Yeah right, I wish.
Little thirteen year old, 100 pound, 5'2 thingymobbersondingalingerdonger me, cannot lift that much.
But I can lift 101:P
Pretty cool word up there, eh?
That's my little creation!
Try saying THAT ONE three times fast! xD
2006-11-27 13:34:03 · answer #4 · answered by Anonymous · 0⤊ 0⤋