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.

 

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

 

Agree with above.

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

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

 

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

 

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??

 

That is an angle of approx 6.5 degrees from horizontal.

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

 

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

 

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.