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# What resistance in pounds is needed to pull this cart up the incline? Answered

Or better yet what would be the formula when the angle of incline, cart weight and hanging weight vary?

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Well, supposing that you pull the rope a small by a small displacement ΔL:

Then the hanging block would rise ΔL vertically, and the rolling block would move upward by (sin(21.8 deg))*ΔL.

So the work it takes to make these blocks move upward against their own weight is :
W = (50 lb)*ΔL + (165 lb)*(sin(21.8 deg))*ΔL

Next I am going to naively assume none of your rope-pulling work is lost to friction, so that:

W = F*ΔL = (50 lb)*ΔL + (165 lb)*(sin(21.8 deg))*ΔL
where F is the force with which you pull on the rope, and with which the rope pulls on you, at that place in the diagram labeled, "?? lbs to pull".

Divide both sides by ΔL, and get :
F = (50 lb) + (165 lb)*(sin(21.8 deg)) = 111.28 lb

Just as an FYI this problem has a real world application. It is a Total Gym (variable resistance via a variable incline using body weight) that is tethered to one side of a Bowflex Machine (resistance provided via bending combinations of carbon-fiber bows). I would be curious as how friction is altering the resistance on the uphill vs. the downhill.

I copied this directly from my old "Riggers Handbook." Notice the problem above doesn't include the coefficient of friction and the load rigged to the pulleys is really designed to merely complicate the problem.

Fp = W h/l
= W sin α

where:

Fp = pulling force (N, lbf)
W = m g = weight of body (N, lbf)
h = elevation (m, ft)
l = length (m, ft)
α = elevation angle (degrees)
m = mass of body (kg, slugs)
g = acceleration of gravity = 9.81 (m/s2) = 32.174 (ft/s2)