Resolve forces into components, add vector forces, and solve for motion along inclines and under angled forces

Learning goal: Find the components of a force at a known angle using trigonometry or a scale diagram; add forces that act at right angles; draw a free-body diagram for an object on a horizontal surface pulled or pushed at an angle and solve for its acceleration, including friction; find the acceleration of an object on a frictionless incline or an incline with a known friction force; and combine the result with kinematic equations to find the time, final velocity or length of the slide.

Forces are vectors. A force has a size and a direction, and when several forces act on an object at different angles, the net force is their vector sum. Two methods give the sum. A scale diagram draws each force as an arrow to scale (for example 1 cm for 10 N) placed tip to tail; the arrow from the start of the first to the tip of the last is the net force, whose length and angle are measured with a ruler and protractor. Trigonometry does the same thing with numbers: each force is split into a horizontal component and a vertical component, the components are added separately, and the net force is rebuilt from the totals. Directions are measured from a stated reference, such as "37° above the horizontal" or as an angle from 0° to 360° measured counter-clockwise from the positive x-axis.

Components. A force F at angle θ above the horizontal has a horizontal component F cos θ and a vertical component F sin θ. Which one gets the cosine depends only on where the angle is measured from: the component along the side adjacent to the angle uses cosine, the opposite side uses sine. A 50 N pull at 37° above the horizontal has F_x = 50 × cos 37° = 50 × 0.799 = 39.9 N and F_y = 50 × sin 37° = 50 × 0.602 = 30.1 N. Check: √(39.9² + 30.1²) = 50 N. The components are not two extra forces; they are the one force described along two directions, and either the force or its pair of components appears on a diagram, never both.

Adding forces at right angles. Forces of 30 N east and 40 N north on a sled combine to a net force of √(30² + 40²) = 50 N at tan⁻¹(40 ÷ 30) = 53° north of east. A scale diagram with 1 cm per 10 N gives a 5.0 cm arrow at the same angle. Forces that are not at right angles are first resolved into components along two perpendicular axes, and then the components are added. For example, a force of 40 N at 30° and a force of 25 N at 135° (both measured counter-clockwise from the positive x-axis) act on one object. Components: 40 cos 30° = 34.6 N and 40 sin 30° = 20.0 N; 25 cos 135° = −17.7 N and 25 sin 135° = 17.7 N. The cosine of an angle between 90° and 180° is negative, which is how the leftward part of the second force enters the sum. Totals: F_x = 34.6 − 17.7 = 16.9 N and F_y = 20.0 + 17.7 = 37.7 N. Net force: √(16.9² + 37.7²) = 41 N at tan⁻¹(37.7 ÷ 16.9) = 66° from the positive x-axis. Because F_x and F_y are both positive, the net force lies between 0° and 90°; when F_x is negative, add 180° to the calculator's angle. A scale diagram, drawn tip to tail, gives the same arrow within the precision of the ruler and protractor.

A fully worked model: a box pulled at an angle. A 10 kg box is pulled along a horizontal floor by a rope at 37° above the horizontal with a tension of 50 N. The floor exerts a kinetic friction force of 12 N (already quantified). Free-body diagram: tension 50 N at 37°, weight 98 N down, normal force N up, friction 12 N backward. Resolve the tension: 39.9 N forward, 30.1 N upward. Vertical direction, no acceleration: N + 30.1 − 98 = 0, so N = 67.9 N. The upward part of the pull has lifted some of the weight off the floor. Horizontal direction: 39.9 − 12 = 10a, so a = 2.8 m/s². If the rope were horizontal the forward component would be the full 50 N and the acceleration 3.8 m/s²; pulling upward at an angle wastes part of the force on lifting, though in a problem where friction were computed from the normal force it would also reduce the friction.

Inclines. For an object on a slope of angle θ, the useful axes are along the slope and perpendicular to it, because the object moves along the slope. The weight mg is then resolved into mg sin θ down the slope and mg cos θ into the slope. Perpendicular to the slope nothing accelerates, so the normal force equals mg cos θ. Along the slope, with no friction, the net force is mg sin θ and the acceleration is a = g sin θ. The mass cancels: on a frictionless incline every object slides down with the same acceleration, just as in free fall, and the incline simply dilutes g by the factor sin θ. At θ = 90° the object falls freely; at θ = 0° it does not move. On a 25° frictionless incline, a = 9.8 × sin 25° = 9.8 × 0.423 = 4.1 m/s². Starting from rest and sliding 3.0 m along the slope: v_f = √(2 × 4.14 × 3.0) = 5.0 m/s and t = √(2 × 3.0 ÷ 4.14) = 1.2 s. A scale diagram gives the same components: draw the weight as an arrow, then drop perpendiculars onto the slope direction and the normal direction and measure.

Inclines with a known friction force. A 4.0 kg block slides down a 30° incline against a friction force of 6.0 N (given; this course does not compute friction on an incline from a coefficient). Along the slope: mg sin 30° = 4.0 × 9.8 × 0.500 = 19.6 N down the slope, friction 6.0 N up the slope, net 13.6 N, a = 13.6 ÷ 4.0 = 3.4 m/s². For a 2.5 m slide from rest, v_f = √(2 × 3.4 × 2.5) = 4.1 m/s and t = √(2 × 2.5 ÷ 3.4) = 1.2 s. With a friction force that is given as a fixed number, the mass no longer cancels: if the same 6.0 N acted on a heavier block, the acceleration would be larger. (Friction that grows in proportion to the mass would cancel again, but calculating friction on an incline from a coefficient is not part of this course.)

Common errors. Using sin where cos belongs: check with a limit, since at θ = 0° the component along a flat floor must be zero, which sin 0° gives. Drawing the normal force vertical on an incline: it is perpendicular to the surface. Adding the weight and its components together on one diagram. Forgetting that a force applied at an angle changes the normal force and, in problems where friction depends on it, the friction too.

Investigation: acceleration down an incline. Materials: a low-friction cart, a track or board at least 1.2 m long, books or a jack to raise one end, a protractor or a meter stick to find the angle from the rise and run, a stopwatch or video, and two extra masses. Steps: set the track to 5°, release the cart from rest at a mark 1.00 m from the bottom, time the run three times, and repeat at 10° and 15°; then at 10° add 0.50 kg and then 1.0 kg to the cart and repeat. Record: angle, mean time, acceleration a = 2d/t², and sin θ. Analyze: plot a against sin θ; the model predicts a straight line of slope g. Predicted accelerations are 0.85, 1.70 and 2.54 m/s², giving times of 1.53, 1.08 and 0.89 s for 1.00 m. A result that supports the claim: accelerations close to g sin θ that do not change when mass is added, because both the gravitational component and the inertia scale with mass. Discuss why the mass has no effect when friction is negligible, and why a friction force that does not grow in proportion to the mass would make the acceleration depend on mass. Safety: catch the cart at the bottom; keep the raised end stable. What the model cannot show: a rolling cart's wheels must spin up, which uses part of the gravitational force and makes the measured acceleration a few percent below g sin θ even with negligible friction; the straight-line plot reveals this as a slope slightly less than 9.8.

Worked example 1 — components of a force at an angle. A rope pulls a crate with 80 N at 60° above the horizontal. Horizontal component: 80 × cos 60° = 80 × 0.500 = 40 N. Vertical component: 80 × sin 60° = 80 × 0.866 = 69 N. Most of this pull lifts rather than drags. The same force at 60° below the horizontal would have the same 40 N forward component but a 69 N downward component that presses the crate into the floor.

Worked example 2 — two perpendicular forces. Two ropes pull a barge: 24 N along the canal and 10 N straight across it. Net force: √(24² + 10²) = √676 = 26 N, at tan⁻¹(10 ÷ 24) = 23° from the canal direction. On a scale diagram with 1 cm per 2 N, the arrows are 12 cm and 5 cm and the resultant measures 13 cm.

Worked example 3 — a frictionless incline with kinematics. A 6.0 kg box is released on a frictionless 20° ramp 4.0 m long. Acceleration: 9.8 × sin 20° = 9.8 × 0.342 = 3.35 m/s². Speed at the bottom: √(2 × 3.35 × 4.0) = 5.2 m/s. Time: √(2 × 4.0 ÷ 3.35) = 1.5 s. Normal force: 6.0 × 9.8 × cos 20° = 58.8 × 0.940 = 55 N, less than the 58.8 N weight because the ramp supports only the perpendicular component.

Worked example 4 — a push at an angle with friction given. A 15 kg box is pushed along a floor by a force of 60 N directed 30° below the horizontal, against a friction force of 20 N. Components of the push: 60 × cos 30° = 52.0 N forward and 60 × sin 30° = 30 N downward. Vertical: N = 147 + 30 = 177 N. Horizontal: 52.0 − 20 = 15a, a = 2.1 m/s². Pushing downward at an angle increases the normal force; in a problem where friction depends on the normal force, that would also increase the friction.

Representation task: a scale diagram and a calculation, side by side. For the 10 kg box pulled at 50 N and 37° with 12 N of friction, draw the free-body diagram with all four forces to scale (1 cm per 10 N), then in a separate sketch resolve the 50 N pull into its components by drawing the right triangle and measuring the legs. Record the measured components beside the calculated ones (39.9 N and 30.1 N), compute the normal force and acceleration from your measured values, and comment on how closely the two methods agree and which is more precise.

Independent performance: incline analysis with a test. Set up an incline at an angle you measure, and either use a low-friction cart or measure the friction force on a sliding block on the incline itself: pull it up the slope at constant speed with a spring scale held parallel to the slope, and subtract the weight component, friction = pull − mg sin θ (friction on the flat track is not the same, because the normal force on the slope is smaller). Predict the acceleration from the free-body diagram, then the time to slide a measured length from rest, then test with three trials. Success criteria: the free-body diagram shows the weight components along and perpendicular to the slope with their computed sizes; the normal force is perpendicular to the surface; the acceleration is derived symbolically before numbers are substituted; the predicted time is compared with the measured time as a percentage; and the discussion explains why, for the frictionless prediction, the mass does not appear.