Weight On Cable

A 50kg truck truck engine is lifted using the setup shown in the image below. An equation of mass mass_engine holds the mass value. That mass is suspended on a wire that is pinned on one end and routed through a pulley on the other. The pulley and pin are at equal heights. The horizontal distance (dist_hor) between the pin and pulley is set to 3meters and the vertical distance (dist_ver) between the top of the weight and the pin and pulley is 1meter.

Solve for the force P in the cable.

d i s t_{v} er =

m

d i s t_{h} or =

m

ma s s_{e} n g in e =

kg

Solution

Assumptions: The pulleys have no friction, so the whole cable carries the same tension, P. The pulley on the engine sits halfway between the pin and the fixed pulley, so the two cable segments are symmetric and each makes the same angle θ with the horizontal.

Step 1: Cable Geometry

Because the setup is symmetric, each cable segment spans half of the horizontal distance and drops the full vertical distance. Together these form a right triangle. The cable's angle from the horizontal comes from the arctangent of the vertical drop divided by the horizontal half-span.

d i s t_{ha l f} = \frac{d i s t _{h or}}{2} = 1.5 m

θ = tan^{- 1} (\frac{d _{v er}}{d _{h or} /2})

θ = arctan (d i s t_{v er} / d i s t_{ha l f}) = 0.588003

Step 2: Free Body Diagram and Sum of Forces

Isolate the pulley attached to the engine. Three forces act on it: the two cable segments, each pulling with tension P at angle θ up and away from the pulley, and the engine's weight W pulling straight down. In the horizontal direction, the two cable forces are equal and opposite, so they cancel. In the vertical direction, the upward parts of both cable forces must hold up the weight.

\sum F_{x} = P cos θ - P cos θ = 0

\sum F_{y} = P sin θ + P sin θ - W = 0 \Rightarrow 2 P sin θ = W

The weight is the engine's mass times gravitational acceleration (the built-in constant grav, 9.81 m/s²).

W = m_{e n g in e} \cdot g

W = ma s s_{e n g in e} \cdot g r a v = 490.5 N

Step 3: Solve for P

Rearrange the vertical force balance to solve for the cable tension:

P = \frac{W}{2 s i n θ}

P = \frac{W}{2 \cdot s i n ( θ )} = 442.131 N

Result: With a 1.5 m half-span and a 1 m drop, θ ≈ 33.7° (0.588 rad), sin θ ≈ 0.555, and W ≈ 490.5 N. This gives a cable force of P ≈ 442 N. P is less than the full weight because two cable segments share the load. But because they pull at a shallow angle, each segment carries more than half of W.