Screw Jack Force Calculator helps calculate lever effort from load, screw lead, mean diameter, thread friction, collar drag, and handle length using torque and efficiency formulas.
How Screw Jack Lifting Force Is Calculated
A screw jack turns rotation at a handle into a much larger linear lifting force. Turning the handle twists the screw against the mating nut thread, and that thread contact plus a rotating collar at the base both resist the motion with friction. The torque you put in at the handle has to overcome both sources of friction before any of it goes toward actually lifting the load.
Torque needed at the screw thread to raise the load:
$$T_{screw} = W \cdot \frac{d_m}{2} \cdot \frac{\mu’ + \tan\lambda}{1 – \mu’ \tan\lambda}$$
where $W$ is the operating load, $d_m$ is the mean screw diameter, and $\tan\lambda = \dfrac{lead}{\pi d_m}$ is the tangent of the thread’s lead angle. $\mu’$ is the effective friction coefficient, which is the entered thread friction $\mu$ adjusted for the thread’s flank angle: $\mu’ = \mu / \cos\beta$, where $\beta$ is half the thread’s included angle. Square threads have no flank angle ($\beta = 0°$, so $\mu’ = \mu$), Acme threads use a 29° included angle ($\beta = 14.5°$), and trapezoidal threads use 30° ($\beta = 15°$).
Torque absorbed by friction at the collar, the flat bearing surface the screw rotates against under load:
$$T_{collar} = W \cdot \frac{d_c}{2} \cdot \mu_c$$
where $d_c$ is the mean collar diameter and $\mu_c$ is the collar friction coefficient. Total raising torque is $T_{raise} = T_{screw} + T_{collar}$, and the force you actually need to apply at the handle is that torque divided by the lever arm: $F = T_{raise} / l$.
Lowering the same load takes a different torque, because the friction and lead-angle terms now work against each other instead of together:
$$T_{lower} = W \cdot \frac{d_m}{2} \cdot \frac{\mu’ – \tan\lambda}{1 + \mu’ \tan\lambda} + T_{collar}$$
If $T_{lower}$ comes out positive, the screw needs torque to lower the load at all, meaning it holds the load in place on its own, self-locking. If it comes out negative, the load can spin the screw backward by itself, and the jack backdrives without a separate brake.
Two more figures come out of the same numbers. Efficiency compares the useful lifting work to the work you put into the handle over one full turn:
$$\eta = \frac{W \cdot lead}{2\pi \cdot T_{raise}}$$
and mechanical advantage is simply $MA = W / F$, how many times the handle force is multiplied into lifting force.
Handle force F acts on lever length l to produce the torque that turns the screw against thread friction and collar friction.
| Symbol | Meaning |
|---|---|
| W | Operational load being lifted |
| lead | Screw lead (linear travel per one full turn) |
| d_m | Mean screw thread diameter |
| mu | Thread friction coefficient (0.01-0.90) |
| d_c | Mean collar diameter |
| mu_c | Collar friction coefficient (0-0.90) |
| l | Handle lever arm length |
| beta | Half the thread’s included angle: 0 deg square, 14.5 deg Acme, 15 deg trapezoidal |
Worked Example
Take a square-thread screw jack raising a 5,000 lb load, with a 0.25 in lead, a 1.5 in mean screw diameter, 0.15 thread friction, a 2.0 in mean collar diameter, 0.10 collar friction, and a 20 in handle lever.
Square threads have no flank angle, so the effective friction stays equal to the entered value: $\mu’ = 0.15$. The lead angle term is $\tan\lambda = 0.25 / (\pi \times 1.5) = 0.0531$.
Thread torque to raise the load: $$T_{screw} = 5{,}000 \times 0.75 \times \dfrac{0.15 + 0.0531}{1 – (0.15)(0.0531)} = 767.55$$ in-lbs.
Collar torque: $T_{collar} = 5{,}000 \times 1.0 \times 0.10 = 500.00$ in-lbs.
Total raising torque: $T_{raise} = 767.55 + 500.00 = 1{,}267.55$ in-lbs, or 105.63 ft-lbs. Handle force: $F = 1{,}267.55 / 20 = 63.38$ lbs.
Efficiency comes out to $$\eta = (5{,}000 \times 0.25) / (2\pi \times 1{,}267.55) = 15.70\%$$, and mechanical advantage is $5{,}000 / 63.38 = 78.89$ to 1.
Checking the lowering direction: $$T_{lower} = 5{,}000 \times 0.75 \times \dfrac{0.15 – 0.0531}{1 + (0.15)(0.0531)} + 500.00 = 860.69$$ in-lbs. That’s positive, so this jack is self-locking, and it would take a 43.03 lb handle force to deliberately lower the load. The lead angle works out to 3.04° and the friction angle to 8.53°, a 5.49° margin, which is why it self-locks: friction angle exceeds lead angle.
| Result | Value |
|---|---|
| Handle force to raise | 63.38 lbs |
| Total raising torque | 105.63 ft-lbs |
| Efficiency | 15.70% |
| Mechanical advantage | 78.89 : 1 |
| Self-locking | Yes |
| Handle force to lower | 43.03 lbs |
| Lead angle / Friction angle | 3.04 deg / 8.53 deg |
What the Result Means
The handle force figure is the raw torque-to-force conversion for one turn of the screw at steady speed. It doesn’t include starting friction, misalignment, or a safety margin, so treat it as the working minimum, not a design load for the handle or gearing.
Efficiency in the 15-20% range, like the worked example above, is normal for a screw jack and isn’t a flaw. Screw jacks trade efficiency for self-locking: the same friction that wastes most of the handle’s work into heat is what keeps the load from dropping when you let go of the handle. A jack with much higher efficiency usually can’t hold a load without a separate brake, which is a bigger safety problem than a stiff handle.
The self-locking status is a direct read of the lowering torque’s sign. Self-locking (positive lowering torque) means the screw needs torque applied to lower the load, so it stays put on its own. Backdrives (negative) means the load will turn the screw and drop under its own weight without a brake or ratchet.
A near-zero result sits right at the boundary between the two and should be treated as unreliable holding, since small changes in friction, from wear, lubrication, or contamination, can flip it either way.
What Changes the Result
Thread profile changes the effective friction coefficient before anything else happens. Square threads keep $\mu’ = \mu$. Acme threads (14.5° half angle) push the effective friction up to about 0.1549 from a 0.15 input, and trapezoidal threads (15° half angle) push it to about 0.1553, roughly 3-3.5% higher than square.
That difference carries through every torque and force output, so an Acme jack always needs slightly more handle force than an otherwise-identical square-thread jack.
Lead and mean screw diameter set the lead angle through $$\tan\lambda = lead / (\pi d_m)$$. A longer lead or a smaller diameter steepens the lead angle, which lowers the raising torque but also pushes the self-locking margin (friction angle minus lead angle) closer to zero.
Push it far enough and the calculator halts entirely: if the effective friction times the lead-angle tangent reaches 1 or more, the raising-torque denominator goes to zero or negative, which is not a physically valid raising condition, and no result is shown.
Collar friction adds a flat, straight-line amount of torque, $W \times d_c/2 \times \mu_c$, on top of the thread torque. It doesn’t touch the lead angle or the self-locking condition at all, since it adds identically to both the raising and lowering torque.
A larger collar diameter or higher collar friction (bearing wear, no thrust bearing, dry contact) raises the handle force needed without changing whether the jack self-locks.
Both friction inputs are bounded: thread friction from 0.01 to 0.90 and collar friction from 0 to 0.90. Values outside that range, a non-positive load, lead, diameter, or lever length, or an efficiency that would come out at 0% or over 100%, all stop the calculation rather than return a number, since none of those describe a real screw jack.
FAQs
What is the formula for the force needed to raise a load with a screw jack?
Handle force equals total raising torque divided by the lever length: $F = T_{raise}/l$. Total raising torque is the thread torque, $W(d_m/2) \times (\mu’+\tan\lambda)/(1-\mu’\tan\lambda)$, plus the collar torque, $W(d_c/2)\mu_c$.
Why does an Acme thread need more handle force than a square thread for the same load?
Acme threads have a 29° included flank angle instead of a square thread’s 0°. That angle divides into the effective friction coefficient as $\mu/\cos(14.5°)$, raising it by roughly 3.3% over the same entered friction value, and that higher effective friction carries straight through to a higher raising torque and handle force.
What does it mean if a screw jack is self-locking?
Self-locking means the lowering torque calculation comes out positive, so the screw needs torque applied to it to lower the load, which is another way of saying it holds the load in place with the handle released. It happens whenever the friction angle is larger than the thread’s lead angle.
What happens if a screw jack backdrives?
Backdriving means the lowering torque comes out negative: the lead angle exceeds the friction angle, so the load’s own weight is enough to spin the screw backward without anyone turning the handle. A jack in that condition needs a separate brake or ratchet to hold the load safely.
How much does collar friction affect the lifting force?
It adds directly and linearly: collar torque is just load times half the collar diameter times the collar friction coefficient. In the worked example above it accounts for 500 of the 1,267.55 in-lbs total raising torque, close to 40% of it, without affecting the lead angle or self-locking outcome at all.
What efficiency should I expect from a screw jack?
Somewhere in the 10-30% range is typical, and 15-20% is common for a lubricated general-purpose jack, like the 15.70% in the worked example. Low efficiency here isn’t wasted design margin, it’s the same friction that lets the jack hold a load without a separate brake.
What friction coefficient range does this calculator accept, and why?
Thread friction is limited to 0.01-0.90 and collar friction to 0-0.90. Those bounds keep the inputs inside physically realistic bearing-surface friction values; a value of zero thread friction would mean no resistance at all, which no real threaded contact has.
Does switching between imperial and metric units change the result?
No, the underlying torque balance is identical either way. Switching units converts the load using 1 lbf = 0.0044482216 kN and every length using 1 in = 25.4 mm, then recalculates from those converted values, so the physical answer stays the same, only the displayed units change (lbs and in-lbs/ft-lbs for imperial, N and N-m for metric).