A family in Lucknow is moving a tall steel almirah, full of clothes, across the room. The son leans on it near the top and pushes. It does not slide; it rocks forward on its front edge, and everyone shouts. His mother kneels and pushes near the bottom instead. The almirah slides across the floor without rocking at all. The push was no larger. Only its height had changed.
Why a body topples
A block resting on the floor feels its weight at its centre of gravity and the floor's forces on its base. The floor can push only where it touches the block, so the normal force must act somewhere within the base. Within those limits it can act at any point. Where exactly it acts is decided by the torques, not by the floor.
Tilt the block a little about one bottom edge, and let go. The weight line, the vertical line through the centre of gravity, still meets the base. The weight's torque about the edge turns the block back, and it settles. Tilt it further, until the weight line passes outside the base. Now no support force can reach under the weight line. The weight's torque about the edge turns the block the other way, and it falls. The block topples.
A body resting on a surface topples when the vertical line through its centre of gravity passes outside its base of support. While that line meets the base, the support forces can balance the weight.
In Fig 5.1 the critical tilt comes when G is straight above E. Then \(\tan\phi = (b/2)/(h/2) = b/h\), where \(\phi\) is the tilt, \(b\) the width and \(h\) the height. For this block \(\tan\phi = 0.6/0.8 = 0.75\), so \(\phi = 36.9°\). A body is harder to topple when \(b\) is large and the centre of gravity is low. That is why racing cars are wide and low, and why heavy goods go at the bottom of a lorry's load.
The toppling rule is the energy picture of the chapter on Potential Energy and Conservation of Energy. As the block tilts about E, its centre of gravity rises until it is straight above E, and then falls. So the upright block sits in a well of potential energy. The balanced position at \(36.9°\) is the top of a hump, an unstable equilibrium. A wider base or a lower centre of gravity makes the well deeper.
A bus can carry heavy luggage on its roof or in a hold below the floor. For safety on a road with many sharp turns, where should the heavy luggage go?
You stand with your feet \(0.3\,\text{m}\) apart, sideways, and lean your whole body sideways without bending. Your centre of gravity is about \(1\,\text{m}\) above the floor. Estimate the greatest lean from the vertical before you must step, in degrees.
Pushing a block at a height
Push a block with a horizontal force \(F\) at a height \(y\) above the floor. Suppose friction holds it still. Then the friction is \(F\), backwards, and the normal force \(N\) equals the weight \(W\). Take torques about the point of the floor under the centre of gravity. The push turns the block forward with torque \(Fy\). Friction acts along the floor, so its moment arm is zero. The weight passes through that point too.
Something must turn the block back. It is the normal force: it moves forward, a distance \(x\) ahead of the centre, so that its torque \(Nx\) balances the push. With \(N = W\), this gives \[ x = \frac{F\,y}{W} \] The higher the push, the further forward the normal force must act. It cannot pass the front edge, at \(b/2\). So the block stays upright only while \(Fy/W \le b/2\). Beyond that the block tips forward about its front edge.
In the sim, \(W = 100\,\text{N}\) and \(F = 40\,\text{N}\), so the normal force acts \(0.4y\) ahead of the centre. At \(y = 0.25\,\text{m}\) it is \(0.10\,\text{m}\) ahead. At \(y = 0.5\,\text{m}\) it reaches the edge, \(0.2\,\text{m}\) ahead. Friction never fails, because it would need more than \(0.5 \times 100 = 50\,\text{N}\). So this block tips at any push height above \(0.5\,\text{m}\), and never slides.
A block pushed harder and harder can fail in two ways, and both must be tested. It tips when \(Fy > Wb/2\), that is when \(F > Wb/(2y)\). It slides when \(F > \mu_s W\), the limiting friction. Whichever limit is smaller is reached first. A push low down makes tipping hard, so the block slides. A push high up makes tipping easy, so the block may tip before friction fails.
A uniform cabinet of mass \(12\,\text{kg}\), \(0.5\,\text{m}\) deep and \(1.5\,\text{m}\) tall, stands on a floor with \(\mu_s = 0.2\). A horizontal push acts at its top edge and slowly grows. Take \(g = 10\,\text{m/s}^2\). Does the cabinet slide or tip first? What if \(\mu_s = 0.15\)?
\(W = 120\,\text{N}\), \(b = 0.5\,\text{m}\) and the push acts at \(y = 1.5\,\text{m}\).
Tip when \(F > Wb/(2y)\). Slide when \(F > \mu_s W\).
Tipping limit: \(120 \times 0.5/(2 \times 1.5) = 20\,\text{N}\). Sliding limit: \(0.2 \times 120 = 24\,\text{N}\). With \(\mu_s = 0.15\), sliding needs only \(0.15 \times 120 = 18\,\text{N}\).
With \(\mu_s = 0.2\) the cabinet tips at \(20\,\text{N}\), before it can slide. On the smoother floor it slides at \(18\,\text{N}\), before it can tip.
A uniform box of mass \(15\,\text{kg}\) has a base \(0.6\,\text{m}\) wide. It stands on a floor with \(\mu_s = 0.5\), and a horizontal push of \(50\,\text{N}\) acts on it. Take \(g = 10\,\text{m/s}^2\). Above what height must the push act for the box to tip, in metres?
On a slope: topple or slide
Put a uniform block on a board and raise one end of the board slowly. The block can fail in two ways. It slides when \(\tan\theta\) passes \(\mu_s\), as the chapter on Friction showed for the angle of repose. It topples when its weight line passes the downhill edge of its base. Tilting the board by \(\theta\) tilts the block by \(\theta\), so it topples when \(\tan\theta\) passes \(b/h\).
Two uniform blocks rest on a board with \(\mu_s = 0.5\). The tall one is \(0.3\,\text{m}\) wide and \(1.2\,\text{m}\) tall; the squat one is \(0.6\,\text{m}\) wide and \(0.75\,\text{m}\) tall. The board is raised slowly. What happens first?
Step through the chips of the figure. At \(10°\) both weight lines meet their bases, and \(\tan 10° = 0.18\) is far below \(\mu_s\). At \(20°\) the tall block's line has passed its downhill edge, so it topples; the squat one stays. At \(30°\), \(\tan 30° = 0.58\) is above \(0.5\), and the squat block slides. A block with \(b/h < \mu_s\) topples first; a block with \(b/h > \mu_s\) slides first.
A uniform block \(0.5\,\text{m}\) wide and \(1.2\,\text{m}\) tall stands on a board with \(\mu_s = 0.6\). The board is raised slowly. What happens first, and at what angle?
Four uniform blocks stand on a floor. Rank them by how far they can be tilted about an edge before they topple, the largest tilt first.
Now return to the almirah. Say it has a mass of \(60\,\text{kg}\), a depth of \(0.6\,\text{m}\) and a height of \(1.8\,\text{m}\), with \(\mu_s = 0.4\). Pushed at the top, it tips once \(F > 600 \times 0.3/1.8 = 100\,\text{N}\). But sliding needs \(0.4 \times 600 = 240\,\text{N}\). A push of \(240\,\text{N}\) avoids tipping only if \(240\,y \le 600 \times 0.3\), that is \(y \le 0.75\,\text{m}\). So the mother pushed below knee height, and it slid.
The bell tower of Pisa, about \(56\,\text{m}\) tall, began to lean while it was being built, because the soil under one side is soft. By 1990 it leaned about \(5.5°\) from the vertical, and the tower was closed. Engineers removed soil from under the higher side, and the tower settled back to a lean of about \(4°\). Its weight line still falls well inside its base, so it stands.
- Assuming the normal force always acts at the middle of the base.It moves within the base to balance the torques. A push high up drives it towards the front edge.
- Testing only for sliding when a block is pushed.Compute the tipping limit \(Wb/(2y)\) and the sliding limit \(\mu_s W\). The smaller one wins.
- Using \(\tan\theta = \mu_s\) for toppling on a slope.Toppling depends on shape, \(\tan\theta = b/h\). Friction only decides sliding.
- Believing a wider base makes a block topple sooner.A wider base keeps the weight line inside for a larger tilt, so the block is harder to topple.