How to Find Tension in a Rope (Single & Connected Masses)

The method: free-body diagram plus Newton's second law

To find tension in a rope, isolate one object, draw a free-body diagram, pick a positive direction, and write Newton's second law, $F_{net} = ma$, along that direction. Tension appears as one force in the equation, and you solve for it. There is no standalone tension formula: $T = mg$, $T = m(g + a)$, and $T = ma$ are all just Newton's second law applied to different situations.

For the setups below, assume a taut, massless, inextensible string with no distributed frictional contact. A rope can only pull, never push, so tension always points along the rope and away from the object. Under this ideal model its tension is uniform; a massless pulley with a frictionless axle redirects it without changing the tension. Inextensibility constrains the motion, while negligible string mass and the ideal pulley assumptions permit the equal-tension equations.

You can practice setting up the diagrams in the free-body diagram builder before working through the examples below.

Hanging mass: T = mg only when a = 0

Assume the hanging mass has only two forces: tension $T$ pulling up and gravity $mg$ pulling down. Taking up as positive, Newton's second law reads

which rearranges to $T = m(g + a)$.

• At rest or moving at constant velocity: $a = 0$, so $T = mg$. A 5.0 kg mass hanging at rest has $T = 5.0 \times 9.8 = 49 \text{ N}$.
• Accelerating upward: $T = m(g + a)$, larger than $mg$. The net force points up; this speeds up an upward-moving mass but slows a downward-moving one.
• Accelerating downward: the signed $a$ is negative, so the same equation $T=m(g+a)$ gives less than $mg$. If $a_d$ denotes the positive downward acceleration magnitude instead, write $T=m(g-a_d)$.

Avoid writing $T = mg$ on reflex. That equality holds only in equilibrium. In this two-force vertical setup, nonzero acceleration means tension differs from weight. Direction of motion alone does not determine tension.

Horizontal pull: vertical forces drop out

For a horizontal rope pulling a block that stays on a level surface, with no other applied forces, split the analysis by direction. Vertically, the normal force balances gravity, so there is no vertical acceleration. Take the pull as positive and let any friction of magnitude $F_f$ oppose that direction:

• Frictionless surface: $T = ma$.
• With friction: $T - F_f = ma$, so $T = ma + F_f$.

Notice that on a frictionless surface the tension does not depend on $g$ at all. A 4.0 kg block accelerating at 1.5 m/s^2 needs only $T = 4.0 \times 1.5 = 6.0 \text{ N}$, far less than its 39.2 N weight, because the rope is not fighting gravity.

If the rope pulls at angle $\theta$ above horizontal, with right and up positive, write $T\cos\theta-F_f=ma_x$ and $N+T\sin\theta-mg=0$ while the block remains on the surface with zero vertical acceleration. Thus $N=mg-T\sin\theta$. If the required normal force is negative, the contact assumption fails. The net force calculator checks $F_{net}=ma$ or perpendicular net components; it does not automatically infer tension, friction, or the normal force from a rope setup.

Two blocks connected by a string: system first, then one block

For two blocks joined by a taut ideal string on the same straight horizontal path, with equal acceleration, use this two-step recipe.

Step 1: Treat both blocks as one system. The string tension is an internal force, so it cancels out of the system equation: the string pulls the back block forward exactly as hard as it pulls the front block backward. Add up only the external forces and divide by the total mass:

Step 2: Isolate one block. Now that you know $a$, draw a free-body diagram of a single block and apply Newton's second law to it alone. Pick the block with the fewest forces on it, usually the trailing block, whose only horizontal force is the tension. Solving that one equation gives $T$.

Trying to pull the tension out of the system equation is the classic dead end: internal forces never appear in it. You need the net force on a single block before $T$ shows up anywhere.

Atwood-style setups: the same recipe over a pulley

An Atwood machine hangs two masses from a string over a pulley; the modified Atwood puts one block on a table instead. Assume one taut, massless, inextensible string over a fixed, massless, frictionless pulley. Choose a separate positive axis for each block following the allowed motion, then add the two scalar force equations. The acceleration magnitudes match, but the acceleration vectors do not point in the same direction.

For a modified Atwood (block of mass $m_1$ on a frictionless table, hanging mass $m_2$), take the table block toward the pulley and the hanging block downward as positive. Adding $T=m_1a$ and $m_2g-T=m_2a$ gives:

For a full Atwood with $m_2 > m_1$, take the heavy side down and the light side up as positive. Adding $m_2g-T=m_2a$ and $T-m_1g=m_1a$ gives:

Then isolate either block to find $T$. A good self-check: for positive unequal masses in this ideal full Atwood setup, $T$ lands strictly between the two weights $m_1 g$ and $m_2 g$. If your answer falls outside that range, recheck the algebra.

Mistakes that cost points

• Assuming $T = mg$ everywhere. For a hanging object with only one vertical tension and weight, this holds when acceleration is zero.
• Doubling the tension because the rope pulls on both ends. If you and a friend each pull the ends of a rope with 100 N, the tension is 100 N, not 200 N. Each person’s pull on the rope pairs with the rope’s pull on that same person. The two forces on the rope’s opposite ends are not a third-law pair. A rope tied to a wall and pulled with 100 N is the identical situation.
• Hunting for tension in the system equation. Internal forces cancel there; you must isolate a single block.
• Keeping a negative tension. Ropes cannot push. First check signs and algebra. If a correct calculation requires a negative tension magnitude, the taut-string model cannot hold. Re-solve with zero tension and check that the resulting motion is consistent with a slack string.
• Giving connected blocks different accelerations. In the single fixed-pulley setups here, a taut, inextensible string constrains the magnitudes to match. Movable pulleys can give different acceleration ratios.

Check tension against net force, then practice

A $2.5\ \mathrm{kg}$ load is supported by one vertical cable and has downward acceleration of magnitude $1.8\ \mathrm{m/s^2}$. Only weight and cable tension act. Take up as positive. Is the tension smaller than, equal to, or greater than the weight? Calculate it using $g=9.8\ \mathrm{m/s^2}$.

Check: $a=-1.8\ \mathrm{m/s^2}$, so $T-mg=ma$ gives $T=2.5(9.8-1.8)=20\ \mathrm{N}$. The weight is $24.5\ \mathrm{N}$ and the net force is $-4.5\ \mathrm{N}$, so tension is smaller than weight.

In the net force calculator, choose solve for F, enter mass=2.5 and acceleration=-1.8. It returns the net force, -4.5 N, not the tension. Recover tension with $T=mg+F_{net}=24.5-4.5=20\ \mathrm{N}$.

Next, open forces and Newton’s laws practice. Start with “Equilibrium: a fixture on a single cable,” then “Two connected blocks on a frictionless table,” then “An Atwood machine.” These move from zero acceleration to system isolation to a fixed-pulley constraint. Draw a separate force diagram before opening each solution.

See OpenStax, Common Forces for the tension model and force diagrams.