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Push a short, wide box and it slides. Push a tall, narrow cupboard near the top and it tips over. Exam questions ask you to predict which happens and at what force, and they combine two chapters that students usually study separately: friction and rigid-body equilibrium. This post gives a two-threshold method, works through a crate and an incline, and lists the traps.
Which syllabus topics does this combine?
The Syllabus for JEE (Main) 2026 lists "static and kinetic friction, laws of friction" under Laws of Motion and "equilibrium of rigid bodies" under Rotational Motion. In NCERT Class 11, static friction is in Laws of Motion, where the law of static friction is fs ≤ μs N, and equilibrium is in Systems of Particles and Rotational Motion, section 6.8. NCERT defines mechanical equilibrium as both the net force and the net torque being zero. Check the syllabus for your exam year.
What are the two thresholds?
Take a uniform block of mass m, base width b and height h on a rough floor with coefficient of static friction μ. Push horizontally with force F at height H above the floor.
Slide threshold. The block slides when F exceeds the limiting friction: F_slide = μ m g.
Topple threshold. As F grows, the normal reaction shifts towards the front edge. At the point of tipping it acts entirely at that edge. Taking torques about the front edge, F × H = m g × (b/2). So F_topple = m g b / (2H).
Whichever threshold is smaller is reached first. Setting them equal shows the block topples before sliding when H > b / (2μ).
Worked example 1: a crate on the floor
A 40 kg crate has a 0.60 m square base and is 1.2 m tall. μ = 0.40 and g = 9.8 m/s². You push horizontally at the top edge (H = 1.2 m).
Weight: m g = 40 × 9.8 = 392 N.
Slide: F_slide = 0.40 × 392 = 156.8 N.
Topple: F_topple = 392 × 0.30 / 1.2 = 98 N.
The crate tips at 98 N before it can slide. Check friction: at 98 N the friction needed is 98 N, below the 156.8 N available, so it does not slide first.
At what height should you push to make it slide instead? Topple first needs H > b / (2μ) = 0.60 / 0.80 = 0.75 m. Push lower than 0.75 m and the crate slides. At H = 0.50 m, F_topple = 392 × 0.30 / 0.50 = 235.2 N, which is above 156.8 N, so it slides at 156.8 N. This is why movers push heavy furniture low.
Worked example 2: a block on a tilting ramp
A uniform block 0.20 m wide (along the slope) and 0.50 m tall stands on a board. μ = 0.60. The board is slowly tilted. Does the block slide or tip first, and at what angle?
Slide when the component of weight down the slope exceeds limiting friction: m g sin θ > μ m g cos θ, so tan θ > μ. That is θ > tan⁻¹ 0.60 ≈ 31.0°.
Tip when the vertical line through the centre of mass passes outside the lower edge of the base. The centre of mass is at height h/2 and half-width b/2 from the edge, so tipping starts when tan θ > (b/2)/(h/2) = b/h = 0.20 / 0.50 = 0.40. That is θ > tan⁻¹ 0.40 ≈ 21.8°.
The block tips at about 21.8°, long before it would slide. A wider block with b/h > μ would slide first.
Why is this harder than it looks?
Physics education research has found persistent difficulties with exactly this kind of reasoning. Ortiz, Heron and Shaffer (2005) gave written questions on balancing to more than 1,000 university students who had completed calculus-based introductory physics. Almost all could handle point-like objects, but when mass was spread out continuously, most students attributed equilibrium to equal forces on both sides of the pivot, rather than equal torques. Many also treated horizontal and tilted bodies at rest as different cases. The authors found these difficulties very persistent, and hands-on activities not based on the research did not help.
For a toppling problem, the lesson is to think in torques about a specific edge, not in "balanced forces".
Which mistakes cost marks?
Taking torques about the centre. You can, but then normal force and friction both enter. Use the tipping edge.
Keeping the normal force at the middle of the base. As the push grows, it moves forward. At tipping, it is at the edge.
Using kinetic friction for the slide threshold. Starting to slide depends on static friction.
Comparing tan θ with h/b instead of b/h on an incline. Check: a very tall, thin block should tip at a small angle.
Forgetting the height of the push. The topple threshold depends on H; the slide threshold does not.
Our posts on friction block problems and torque and ladder equilibrium cover each half of this topic in more depth.
A practice routine
For every block problem, write both thresholds before deciding anything.
Vary one quantity at a time: push height, μ, then b/h, and predict which changes the answer.
Do two incline problems: one with b/h > μ and one with b/h < μ.
Sketch the normal force moving to the edge in each diagram.
If torque about an edge still feels abstract, the Fermi Concept Library covers every concept in physics with its own lesson, so you can revise torque and equilibrium before returning to these problems.
Find a tall box at home, push it gently at different heights, and predict before each push whether it will slide or tip.
