Resultant force from a three-way tug-of-war vector diagram

Key concept: Vector addition for a three-way tug-of-war determines both the resultant force and its direction on the knot.

Step by step

Set up the vector picture

In a three-way tug-of-war, the knot experiences three forces at once: 210 N, 190 N, and 200 N. Since the ropes make equal angles around the knot, the forces should be treated as vectors rather than simple scalars. The drawing in part (a) is not just a sketch; it is the tool that shows how the directions combine.

For a scale diagram, choose a convenient scale such as 1 cm = 20 N. Then draw three arrows from a common point with lengths 10.5 cm, 9.5 cm, and 10.0 cm, placing them at the equal angles shown by the problem. The resultant is the single arrow from the tail of the first to the head of the last when the vectors are arranged tip-to-tail.

Find the resultant by components or a scale diagram

Because the angles between the ropes are equal, the directions are spaced evenly around the knot. A convenient analytical approach is to place the three forces head-to-tail and use vector addition. The resultant is the vector sum of all three forces.

If the geometry forms equal 120° separations, then the components can be written using standard x- and y-axes. One force can be chosen along the positive x-axis, the second at 120°, and the third at 240°. Then add the x-components and y-components separately.

The magnitude of the resultant is the size of the final vector after component addition. The direction is the angle measured from the chosen reference axis. Since the forces are not equal, the forces will not cancel perfectly, so the resultant will point toward the side with the larger effective pull.[1][2]

Why the answer is not zero

If all three pulls were the same size, equal spacing could produce a balanced knot. Here the sizes are 210 N, 190 N, and 200 N, so the imbalance is small but real. The difference between the pulls creates a nonzero resultant, even though the angles are symmetric.

That is why a neat vector diagram is so useful: it makes the imbalance visible before any calculation is done. The final answer should be checked against the sketch. If the strongest force is 210 N and the weakest is 190 N, the resultant should be much smaller than 600 N and should point closer to the 210 N rope than to the 190 N rope.

Common interpretation check

The phrase “angles between the ropes are equal” means the directions are evenly spaced around the knot, not that the force magnitudes are equal. That distinction matters because vector problems depend on both size and direction. In a scale diagram, equal spacing of angles with unequal arrow lengths leads to a short but definite resultant. A careful component method or a well-drawn scale diagram should give the same conclusion.[1][2]

Pitfall alert

A common mistake in this three-rope knot problem is adding 210 N + 190 N + 200 N as if the forces were all in the same direction. Forces with different directions do not combine like ordinary numbers. Another trap is drawing the three arrows end-to-end but forgetting that the direction of the resultant comes from the closing side of the vector polygon, not from the longest individual rope. Students also sometimes place the ropes at equal angles but forget that the actual angle value matters; if the spacing is wrong, the component sums are wrong even when the lengths are correct. Finally, the scale diagram must be labeled carefully, because a small error in scale or angle can produce a noticeably different resultant direction. Always compare the diagram with the expected symmetry before reading off the final vector.

Try different conditions

If Allen pulled with 240 N instead of 200 N while Niki remained at 210 N and Jeanette at 190 N, the vector balance would shift toward Allen’s rope and the resultant would become larger in that direction. If the equal spacing between ropes were changed so the ropes were 90° apart instead of evenly spaced around the knot, the same three forces would need to be resolved with a different component layout, and the resultant magnitude could change substantially. A useful variation is to keep the angles equal but change the forces to 200 N, 200 N, and 200 N; in that case the vector sum would be zero if the directions are evenly spaced by 120°. That transformed version helps show that equal magnitudes and symmetric angles can produce equilibrium, while unequal magnitudes produce a net pull.

Further reading

vector addition of forces, resultant force direction, scale diagram vectors