mj

Ok, I have to move a machine weighing 5T down a ramp and contol the weight with a turfor ( winch)

If the machine was dangling at the end of the rope at 90 degrees, then obviousley there will be 5T on the rope, if the machine was on a ramp at 45 degrees would there be 2 .5T on the rope??.

what about on a 15 degree slope?

my graph says just over 850 kgs @ 15 degrees. is this about right??

not taking into acount any friction for rollers/skates etc, am I correct??

TIA

mike.

ajm

If I have understood the problem correctly, the ramp is at 15 degrees to the horizontal so the component of the weight acting down the ramp is 5,000 x sin(15) = 1294kg

matt.bowey

Agree with above.

Force on Rope = Mass x Sin theta = 1.29T.

so 45 deg is 0.707 of 5T not half...!

Dr Steve

ajm and matt are correct assuming rope is pulling parallel to the ramp. If you are pulling horizontal then force of 1340kg = 5000xtan(15)

mj

I have looked more cloeseley at the drawing, the ramp is 15 m long and about 1.7m high.

I off off that day at school when we did trig, how does that work out??

Dr Steve

That is an angle of approx 6.5 degrees from horizontal.

Force is: 5000 x sin(6.5) = 567kg

ajm

<pedant>p.s. if you are using the word "force" you should also be multiplying by g and quoting in Newtons!</pedant>

Daz34

Maximum load for a standard Tirfor (Model T516) winch is 1600kg (the limit of the wire rope). 1500kg if you are using the D shackles.