Equilibrium, Sliding and Tipping | JEE Notes

Statics is a constraint problem. You must determine whether the available contact forces can satisfy both force balance and torque balance.

\sum 𝐅 = 0, \sum τ_{O} = 0 .

When the first equation holds, the second may be evaluated about any point. Choose a point through which several unknown forces act; they then disappear from that torque equation.

Start with a free-body diagram of one isolated body. Replace every contact with its force: a smooth wall supplies a normal force; a rough floor supplies normal and friction components; a light taut string pulls along itself. Static friction satisfies $| f | \leq μ_{s} N$ . Equality applies at impending slip, not at every equilibrium.

Contact also has a geometric restriction: an ordinary surface can push but cannot pull. A calculated negative normal force means the assumed contact cannot persist. For a block supported over a finite base, the resultant normal force can move within that base. Tipping starts when it reaches an edge, beyond which equilibrium would require an impossible pressure distribution.

Checkpoint: how do you choose the direction of unknown static friction?

You can infer the likely slipping tendency and draw friction against it, or choose an arbitrary direction and solve. A negative answer simply reverses the chosen direction. What matters is checking the final magnitude against $μ_{s} N$ . Friction opposes the tendency of relative slipping at the contact, so it need not oppose the centre's motion. In an equilibrium problem the centre does not move at all, yet friction can be nonzero.

Find the friction actually needed

A uniform ladder of length $5 m$ and weight $200 N$ rests against a smooth vertical wall. Its foot is $3 m$ from the wall and its top is $4 m$ above the floor. A person of weight $600 N$ stands $4 m$ along the ladder from its foot. Find all contact-force components and the minimum floor coefficient of static friction.

Force diagram is schematic; use the stated 3–4–5 geometry for all distances and moments.

Hint: take moments about the foot

The wall force acts horizontally at a height of $4 m$ . Each weight's torque arm is its horizontal distance from the foot.

Full solution

The ladder has $\cos θ = 3 / 5$ . The centre's horizontal distance from the foot is $(2.5) (3 / 5) = 1.5 m$ ; the person's is $(4) (3 / 5) = 2.4 m$ . Torque balance about the foot gives

4 N_{w a l l} = 200 (1.5) + 600 (2.4), N_{w a l l} = 435 N .

The wall pushes the ladder away from itself, so floor friction acts towards the wall. Force balance gives

f = 435 N, N = 200 + 600 = 800 N .

For equilibrium, $435 \leq μ_{s} (800)$ , so $μ_{s, \min} = 0.54375$ . If $μ_{s} = 0.70$ , the actual friction is still $435 N$ , not $560 N$ .

Check: Moving the person towards the foot reduces the clockwise weight torque and therefore the required friction. This matches physical intuition.

Which failure happens first?

A rectangular block weighs $200 N$ , has base width $0.30 m$ in the push direction, and rests on a horizontal floor with $μ_{s} = 0.40$ . A horizontal force $P$ acts at height $0.60 m$ . The centre of mass lies over the base midpoint. As $P$ increases slowly, does the block slide or tip first?

Watch the animation: Topple or slide?

Try the simulator: Slide or topple?

Hint: calculate two thresholds under the equilibrium assumption

At tipping, take moments about the forward bottom edge. At sliding, use the limiting floor friction.

Full solution

Vertical balance gives $N = W = 200 N$ , so the sliding threshold would be $P_{s l i d e} = μ_{s} W = 80 N$ . At impending tipping, the normal reaction passes through the forward edge, while the weight has lever arm $b / 2$ :

P_{t i p} h = W \frac{b}{2}, P_{t i p} = \frac{200 (0.15)}{0.60} = 50 N .

Thus the block tips first. At this threshold, friction need only be $50 N$ , below the available $80 N$ , so the conclusion is self-consistent. Before tipping, the normal resultant shifts forward by $x = P h / W$ ; it reaches $b / 2$ at $P = 50 N$ .

Check: Tipping precedes sliding when $b / (2 h) < μ_{s}$ . Lowering the push reduces its overturning effect. These thresholds describe the onset, not the subsequent motion.

Toppling on a slope

A block of base $b$ and height $h$ on a tilting board slides when $\tan θ = μ$ . It topples when the weight line passes the lower edge, at $\tan θ = b / h$ . So it topples first if $b / h < μ$ . Tall, narrow blocks on rough boards topple; short, wide blocks on smooth boards slide.

Equilibrium, friction and the first loss of support

In 60 seconds

Find the friction actually needed

Which failure happens first?

Toppling on a slope