A free body diagram strips a problem down to one object and the external forces acting on it, and the force that decides most of the rest is the normal force: N = mg cosθ − F sinφ. Move the mass, surface angle, applied force and friction sliders below and watch every arrow resize and rotate to match.
Tilt the surface, load the block, pull on it at an angle and change the grip. Every arrow is drawn to its true size on one shared scale, so their lengths can be compared directly. Watch what the normal force does as the slope steepens — it does not follow the weight.
Begin with the two sliders that describe the situation rather than the forces. Mass sets how heavy the block is, and moving it scales the weight arrow, the normal force and the friction available all at once. Surface angle tilts the whole scene, and this is the slider worth dragging slowly, because two readouts move against each other as it does: the down-slope share of the weight climbs while the normal force falls away. That trade is the reason a ramp gets harder to hold something on twice over, and the geometry behind it is set out in the guide to inclined plane physics.
The next pair describes what you are doing to the block. Applied force pushes it along the surface; pull angle lifts the line of that push up off the surface, from flat along it to straight away from it. Raise the pull angle and the normal force drops even on a level floor, because part of the pull is now carrying the block rather than driving it. Push far enough and the normal force hits zero, the status line reports that the block has left the surface, and friction vanishes at the same instant — there is no contact left to produce any.
The last two sliders are the grip. Static friction sets the ceiling on how much the surface can hold before anything moves, and kinetic friction sets what it supplies once the block is sliding. While the block is held, the friction readout shows the force actually in play rather than that ceiling, which is the point most force problems are really testing; the difference between those two numbers is spelled out in the guide to what friction is and how it works. Drop both coefficients to zero and the friction arrow disappears entirely, leaving the cleanest possible three-arrow diagram.
Read the picture as a picture. Every arrow is drawn on one shared scale, so a shaft twice as long really is a force twice as large and you can compare any two of them by eye. The dashed right-angled triangle is the resolution step done for you: its two legs are the weight split along and into the surface, and they land exactly on the tip of the weight arrow, because that is what resolving a vector means. When you want the net force turned into a number rather than a shape, the Newton's Second Law calculator runs the same ΣF = ma step with the working shown.
Press Worked example: pushed crate to load the first row, then move a single slider per step. Every figure in the table is the string the simulator itself printed for those slider positions — the values were read back out of the running lab, not worked out on paper — so if the table and the tool ever disagree, the tool is right. Both coefficients stay at 0.25 throughout.
| Step | Normal force N | Weight down the slope | Friction f | Acceleration a | Status |
|---|---|---|---|---|---|
| Reset: 12 kg on the flat, pushed with 45 N | 117.72 N | 0 N | 29.43 N | 1.2975 m/s² | Sliding and speeding up |
| Tilt the surface to 20 degrees | 110.62 N | 40.26 N | 27.66 N | 4.8006 m/s² | Sliding and speeding up |
| Tilt it further, to 40 degrees | 90.18 N | 75.67 N | 22.54 N | 8.1770 m/s² | Sliding and speeding up |
| Back to flat; raise the pull angle to 30 degrees | 95.22 N | 0 N | 23.80 N | 1.2638 m/s² | Sliding and speeding up |
| Back to a level pull; double the mass to 24 kg | 235.44 N | 0 N | 45.00 N | 0 m/s² | Held at rest |
| 12 kg again, no push, on a 20 degree slope | 110.62 N | 40.26 N | 27.66 N | 1.0506 m/s² | Sliding and speeding up |
The first three rows are the misconception this simulator exists to correct. The block never changes and its weight never changes, yet the normal force falls from 117.72 N to 90.18 N as the surface tilts, and the friction available falls with it — at exactly the moment the down-slope pull has grown from nothing to 75.67 N. Row four reaches a similar N by a completely different route, with the surface flat and the pull angled upward instead. Row five is the static case: doubling the mass raises the ceiling on friction above the 45 N push, so friction supplies exactly 45.00 N, the acceleration is zero and the block does not move at all.
Because the normal force only ever has to balance the part of the weight that presses into the surface, and tilting the surface makes that part smaller. Resolve the weight onto axes that follow the slope and it splits into mg sin(theta) along the surface and mg cos(theta) into it; only the second one is the surface's problem, so N = mg cos(theta). At 0 degrees the cosine is 1 and N carries the whole weight. By 40 degrees it is down to about 0.766 of it, and the down-slope piece has grown from nothing to two thirds of the weight. Watch the two readouts move in opposite directions as you drag the angle slider and the whole idea lands in about five seconds.
The friction readout goes to zero and stays there, and the block accelerates on the smallest tilt you can set. With both coefficients at zero there is nothing to resist the down-slope pull, so the acceleration becomes the applied force divided by the mass plus g sin(theta), and on a slope with no push at all it is exactly g sin(theta). It is worth doing this once with the applied force at zero, because it isolates gravity's contribution: every number the panel shows then comes from the tilt alone, and you can check it against g times the sine of the angle on a calculator.
Because static friction supplies whatever force is needed to prevent sliding, up to a ceiling, and below that ceiling nothing moves. The status line reads Held at rest and the friction readout shows the force actually being used, not the maximum available, which is the distinction most force problems turn on. Keep raising the angle and you will find the exact degree at which it breaks away: it is the angle whose tangent equals the static coefficient, so at mu-s = 0.25 the block holds up to about 14 degrees and slips past it. Set the applied force to zero first, or the push moves the threshold.
No, and the simulator makes that easy to prove. Set both friction coefficients and the applied force to zero, note the acceleration, then drag the mass from 0.5 kg to 100 kg. The weight readout climbs by a factor of two hundred and so does the normal force, but the acceleration does not move at all. Gravity's pull scales with mass and the resistance to being accelerated scales with mass in exactly the same proportion, so the two cancel and every object slides at g sin(theta). Add any friction or any push and mass starts to matter again, because those forces do not scale with it.
Because a pull angled up off the surface is partly lifting the block, and the surface only has to support what is left. The lifting part is F sin(phi), which comes straight out of the normal force: N = mg cos(theta) minus F sin(phi). Raise the pull angle and you can watch N fall while the weight readout does not move at all, which is the cleanest demonstration on the page that N is not a property of the object. Keep going and N reaches zero, the status line reads Off the surface, and friction disappears with it, because there is no contact left to generate any.