Force Components & Equilibrium
A short lesson with diagrams, a worked example, a problem to try, and a quick check. Then quiz yourself on 4 practice questions.
Break every angled force into horizontal and vertical components. An object is in equilibrium only when the forces cancel in each direction separately: ΣFx = 0 and ΣFy = 0.
1. Splitting an angled force
2. Balancing each direction
For the sled moving at constant velocity:
Horizontal: FT cos θ = Ff
Vertical: Fn + FT sin θ = mg, so Fn = mg − FT sin θ
Two forces at right angles? Draw them to scale as the sides of a rectangle. The diagonal is their resultant: it points between them, closer to the larger force.
Check your sines and cosines with an extreme case: if θ = 0 (rope flat), the vertical part F sin 0 must be zero. ✓
3. Example, then try it
A 10 kg sled is pulled at constant velocity by a 50 N rope at 37° above horizontal. Find the friction force and the normal force. (cos 37° ≈ 0.80, sin 37° ≈ 0.60)
Horizontal: Ff = 50(0.80) = 40 N
Vertical: Fn = mg − FT sin θ = 100 − 50(0.60) = 70 N, less than the 100 N weight.
A 60 N sign hangs from two cables. Each cable makes a 60° angle with the horizontal, one on each side. Find the tension in each cable. (sin 60° ≈ 0.87)
Horizontal: the two cables pull sideways equally in opposite directions, so ΣFx = 0 automatically.
Vertical: each cable pulls up with FT sin 60°, so 2FT(0.87) = 60 N.
Solve: FT = 60 ÷ 1.74 ≈ 35 N.
Notice: each cable pulls harder than half the weight (30 N), because only part of its pull is upward.
Answer: about 35 N in each cable. The flatter the cables, the larger the tension gets.
4. Common mistakes
A box is pulled across the floor by a rope angled 30° above horizontal. Compared with pulling it with a horizontal rope, the normal force on the box is…
The rope’s pull has an upward component that helps hold the box up, so the floor pushes less: Fn = mg − FT sin 30°.
Ready to practice?
4 questions on Force Components & Equilibrium, each with a hint and an explanation for every wrong answer.