Classical-Mechanics · Unit 9 · Video 2 · Interactive Practice

Two Bodies, One String: Solving the Cart-and-Pulley Problem

IKey Formulas

Formula	Name	Where it comes from
$T - \mu_k m_C\, g = m_C\, a$	Cart: Newton's 2nd law	Horizontal, using $N = m_C g$
$m_B\, g - T = m_B\, a$	Block: Newton's 2nd law	Down chosen positive
$a = \dfrac{m_B\, g - \mu_k m_C\, g}{m_C + m_B}$	Acceleration	Solve the two together
$T = (\mu_k + 1)\,\dfrac{m_C\, m_B}{m_C + m_B}\,g$	String tension	Back-substitute $a$

Key Insight: Two constraints fuse the free bodies — a massless rope gives one tension $T$ , an inextensible string gives one acceleration $a$ — collapsing four force equations into two.

IIVisualization 1 — The Two Free Bodies

Each body's forces resolve on its own axes, and the net force along its motion equals $ma$ .

IIIVisualization 2 — One String, One Acceleration

An inextensible string makes the cart and block travel equal distances, so they share a single acceleration.

Time

💡 The rope's other job: being massless, it carries one tension end to end, $T_{R,C} = T_{R,B} = T$ , even while both bodies accelerate.

IVVisualization 3 — Friction vs. the Frictionless Limit

With friction on, the acceleration falls short of the frictionless benchmark $a_0 = \dfrac{m_B g}{m_C + m_B}$ .

Your prediction:

a

2.0

💡 If $\mu_k \ge \dfrac{m_B}{m_C}$ the friction wins — the block's weight cannot overcome it, $a \le 0$ , and nothing moves.

VQuiz Questions

Problem 1 · Kinetic Friction on the Cart

Given: cart mass $m_C = 2\ \text{kg}$ , $\mu_k = 0.25$ , $g = 9.8\ \text{m/s}^2$ — find the kinetic friction force $f_k$ acting on the cart.

✅ Correct!

f_k = \mu_k N = \mu_k m_C g = 0.25(2)(9.8) = 4.9\ \text{N}

❌ Close. That is the normal force

N = m_C g = 19.6\ \text{N}

; friction is

\mu_k

times it.

❌ Not quite. You dropped

g

f_k = \mu_k m_C g

, not

\mu_k m_C

❌ Not quite. You dropped the mass:

f_k = \mu_k m_C g

, not

\mu_k g

❌ Not quite. Use all three factors:

f_k = \mu_k m_C g

Show solution

The vertical equation gives the normal force:

N = m_C g = (2)(9.8) = 19.6\ \text{N}

Kinetic friction is $\mu_k$ times the normal force:

f_k = \mu_k N = (0.25)(19.6) = 4.9\ \text{N}

Problem 2 · Acceleration — Don't Forget Friction

Given: $m_C = 3\ \text{kg}$ , $m_B = 2\ \text{kg}$ , $\mu_k = 0.1$ , $g = 10\ \text{m/s}^2$ — find the acceleration $a$ .

✅ Correct!

a = \dfrac{m_B g - \mu_k m_C g}{m_C + m_B} = \dfrac{20 - 3}{5} = 3.4\ \text{m/s}^2

❌ Close.

4.0

is the frictionless value

\frac{m_B g}{m_C+m_B}

. Subtract the friction term

\mu_k m_C g = 3\ \text{N}

from the numerator.

❌ Not quite. The block's weight drives the system; the numerator is

m_B g - \mu_k m_C g

, not

m_C g - \mu_k m_C g

❌ Not quite. Divide by the total mass

m_C + m_B = 5\ \text{kg}

, not by

m_B

alone.

❌ Not quite.

a = \frac{m_B g - \mu_k m_C g}{m_C+m_B}

Show solution

The block's weight drives the motion; friction on the cart resists it:

a = \frac{m_B g - \mu_k m_C g}{m_C + m_B} = \frac{(2)(10) - (0.1)(3)(10)}{3 + 2} = \frac{20 - 3}{5} = 3.4\ \text{m/s}^2

Problem 3 · Acceleration, Then Tension

Given: $m_C = 1\ \text{kg}$ , $m_B = 1\ \text{kg}$ , $\mu_k = 0.5$ , $g = 10\ \text{m/s}^2$ — find the acceleration $a$ , then the tension $T$ .

What is the acceleration $a$ ?

What is the tension $T$ ?

✅ Correct!

a = \frac{10 - 5}{2} = 2.5\ \text{m/s}^2

and

T = m_B g - m_B a = 10 - 2.5 = 7.5\ \text{N}

❌ Check the acceleration.

a = \frac{m_B g - \mu_k m_C g}{m_C+m_B} = \frac{10 - 5}{2}

; the value

5.0

forgets the friction term

\mu_k m_C g = 5\ \text{N}

❌ Check the tension. The block accelerates, so

T \ne m_B g

. Use

T = m_B g - m_B a = 10 - 2.5

Show solution

Step 1 — acceleration:

a = \frac{m_B g - \mu_k m_C g}{m_C + m_B} = \frac{(1)(10) - (0.5)(1)(10)}{1 + 1} = \frac{10 - 5}{2} = 2.5\ \text{m/s}^2

Step 2 — tension (from the block equation $m_B g - T = m_B a$ ):

T = m_B g - m_B a = (1)(10) - (1)(2.5) = 7.5\ \text{N}

Check with the tension formula: $T = (\mu_k + 1)\frac{m_C m_B}{m_C+m_B}g = (1.5)\frac{1}{2}(10) = 7.5\ \text{N}$ ✓

Problem 4 · Double Both Masses

Given: a working cart-and-pulley system. If both $m_C$ and $m_B$ are doubled while $\mu_k$ and $g$ stay fixed, how do the acceleration $a$ and the tension $T$ change?

✅ Correct! In

a = \frac{(m_B - \mu_k m_C)g}{m_C+m_B}

top and bottom both scale by

2

, so

a

is unchanged; in

T = (\mu_k+1)\frac{m_C m_B}{m_C+m_B}g

the factor

\frac{m_C m_B}{m_C+m_B}

scales by

2

, so

T

doubles.

❌ Not quite. Only

T

doubles — in

a

, numerator and denominator scale together and cancel.

❌ Not quite.

a

is unchanged, but

\frac{m_C m_B}{m_C+m_B}

has degree

1

in the masses, so

T

doubles.

❌ Not quite.

a

does not shrink — its numerator and denominator both scale by

2

and cancel.

❌ Not quite. Compare degrees:

a

is degree

0

in the masses,

T

is degree

1

Show solution

Replace $m_C \to 2m_C$ and $m_B \to 2m_B$ .

Acceleration:

a = \frac{2m_B g - \mu_k (2m_C) g}{2m_C + 2m_B} = \frac{2(m_B g - \mu_k m_C g)}{2(m_C + m_B)} = \frac{m_B g - \mu_k m_C g}{m_C + m_B}

unchanged. Tension:

T = (\mu_k+1)\frac{(2m_C)(2m_B)}{2m_C + 2m_B}g = (\mu_k+1)\frac{4 m_C m_B}{2(m_C+m_B)}g = 2\,(\mu_k+1)\frac{m_C m_B}{m_C+m_B}g

exactly double.

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