Newton's Second Law in Two Dimensions
Resolve the forces along the direction of motion and perpendicular to it, then apply F = ma along that direction while the perpendicular components balance. This is what makes a slope problem solvable.
What a learner can do afterwards
- Find the acceleration of a block sliding down a smooth slope
- Resolve perpendicular to the slope to find the normal reaction
- Handle a pulling force applied at an angle to the direction of travel
1 · Read
Tilt your axes along and across the slope: motion happens along the surface. Weight mg splits into mg sin(theta) pulling down the slope and mg cos(theta) pressing into it. For 30 degrees, sin30 = 0.5, so the downslope pull is half the weight.
On a smooth slope nothing else pushes, so F = ma along the surface gives a = g sin(theta), the same for heavy and light blocks. At 30 degrees with g = 9.8, a = 9.8 times 0.5 = 4.9 m/s squared. Across the slope there is no motion, so the normal reaction balances mg cos(theta) and stays below mg.
A rope at an angle resolves the same way: a 20 N pull at 60 degrees to travel gives 20 cos60 = 10 N forward, since cos60 = 0.5. Its upward piece carries some weight and lightens the normal reaction. Forward drives, upward unloads: two effects from one force.
Steeper means faster: g sin(theta) grows toward g as the slope steepens, but stays below g for any genuine slope. Against a 30 N pull uphill, a 4 kg block feels 4 times 9.8 times 0.5 = 19.6 N downhill, leaving 30 - 19.6 = 10.4 N up the slope.
Resolve weight along and across the slope, then run F = ma down the hill and balance across it.
2 · Watch
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Where it sits
8 questions wait behind this lesson, each with its answer explained. Every answer feeds the sky: stars light as they are learned, and dim when it is time to come back.