A porter at a railway station holds a loaded handcart on the ramp to the platform. If the rope goes slack, the cart creeps down. If the porter pulls very hard, the cart starts up the ramp. Between these two, there is a wide band of pulls for which the cart does not move at all.
A band of pulls, not one pull
On a smooth incline, only one pull along the slope holds a block: \(P = mg\sin\theta\), exactly the component of the weight down the slope. On a rough incline, friction widens this single value into a band. If the pull is a little less than \(mg\sin\theta\), friction helps it by pointing up the slope. If the pull is a little more, friction resists it by pointing down the slope. In both cases the block stays.
Take up the slope as positive, and let \(f\) be the friction up the slope. For a block at rest, \(P + f - mg\sin\theta = 0\), so \(f = mg\sin\theta - P\). The perpendicular balance gives \(N = mg\cos\theta\), so the size of \(f\) can be at most \(\mu_s mg\cos\theta\). The block stays as long as \(|mg\sin\theta - P| \le \mu_s mg\cos\theta\).
A \(10\,\text{kg}\) block rests on a \(37°\) incline with \(\mu_s = 0.5\). It is held by a pull of \(60\,\text{N}\) up the slope. The component of its weight down the slope is \(mg\sin 37° = 60\,\text{N}\). Which way does friction point?
The two ends of the band
Step through the chips of Fig 2.1. The limit here is \(\mu_s mg\cos 37° = 40\,\text{N}\). At \(40\,\text{N}\) of pull, friction is \(20\,\text{N}\) up the slope. At \(20\,\text{N}\) it is \(40\,\text{N}\) up the slope, at its limit. At \(100\,\text{N}\) it is \(40\,\text{N}\) down the slope, at its limit again.
Walk the pull down from \(60\,\text{N}\) in Fig 2.1. Friction grows up the slope until it reaches its limit. That is the least pull, \(P_{\min}\). Any weaker pull and the block slides down. Now walk the pull up from \(60\,\text{N}\). Friction grows down the slope until it reaches its limit again. That is the greatest pull, \(P_{\max}\). Any stronger pull and the block slides up.
\[ N = mg\cos\theta \]
Why. nothing moves perpendicular to the slope, so the normal force balances the perpendicular component of the weight.
\[ P_{\min} + \mu_s N - mg\sin\theta = 0 \]
Why. at the least pull the block is about to slide down, so friction is at its limit and points up the slope, helping the pull.
\[ P_{\min} = mg(\sin\theta - \mu_s\cos\theta) \]
Read it. put \(N = mg\cos\theta\) and solve for the pull.
\[ N = mg\cos\theta \]
Why. the perpendicular balance is the same as before.
?
Your turn. which equation holds when the block is about to slide up?
\[ P_{\max} = mg(\sin\theta + \mu_s\cos\theta) \]
Read it. put \(N = mg\cos\theta\) and solve for the pull.
The width of the band is \(P_{\max} - P_{\min} = 2\mu_s mg\cos\theta\). It is twice the limiting friction, because friction can swing from its full value up the slope to its full value down the slope. A rougher incline gives a wider band. On a smooth incline the band closes to the single value \(mg\sin\theta\).
A \(4\,\text{kg}\) block rests on a \(37°\) incline with \(\mu_s = 0.25\). Take \(g = 10\,\text{m/s}^2\), \(\sin 37° = 0.6\), \(\cos 37° = 0.8\). What is the greatest pull along the slope, in newtons, for which the block stays at rest?
For the block of Fig 2.1, for which pulls does friction point down the slope while the block stays at rest?
A loaded handcart of total mass \(50\,\text{kg}\) stands on a ramp at \(37°\). Treat it as a block with \(\mu_s = 0.5\) against the ramp. Take \(g = 10\,\text{m/s}^2\), \(\sin 37° = 0.6\), \(\cos 37° = 0.8\). Find the least and the greatest pull along the ramp that hold it.
Weight component down the ramp: \(500 \times 0.6 = 300\,\text{N}\). Normal force: \(500 \times 0.8 = 400\,\text{N}\). Limiting friction: \(0.5 \times 400 = 200\,\text{N}\).
\(P_{\min} = mg\sin\theta - \mu_s N\) and \(P_{\max} = mg\sin\theta + \mu_s N\).
\(P_{\min} = 300 - 200 = 100\,\text{N}\). \(P_{\max} = 300 + 200 = 500\,\text{N}\).
Any pull from \(100\,\text{N}\) to \(500\,\text{N}\) holds the cart. The porter uses the least effort at \(100\,\text{N}\), where friction supplies two thirds of the holding force.
A block that rests by itself
If \(\tan\theta \le \mu_s\), the incline is below the angle of repose and the block rests with no pull at all. Friction then points up the slope and equals \(mg\sin\theta\). To start this block moving, you may pull it up the slope or down the slope. The friction flips each time, because it always opposes the way the block is about to move.
The two pulls in Fig 2.2 are very unequal. Pulling up, you fight both the weight component and the limiting friction: \(30 + 36 = 66\,\text{N}\). Pulling down, the weight component is on your side. You only supply what friction can hold beyond it: \(36 - 30 = 6\,\text{N}\).
A \(2\,\text{kg}\) block rests by itself on a \(37°\) incline with \(\mu_s = 1\). Take \(g = 10\,\text{m/s}^2\), \(\sin 37° = 0.6\), \(\cos 37° = 0.8\). What pull down the slope, in newtons, just starts it sliding down?
For the \(5\,\text{kg}\) block of Fig 2.2, rank these pulls from the smallest to the largest.
A truck parked on a steep ghat road in the Western Ghats is held by friction at its braked wheels, pointing up the road. The driver leaves it in gear and puts stones behind the wheels as well. The stones widen the band of safe conditions. A heavy load or a wet road lowers the limit \(\mu_s N\). When the limit falls below \(mg\sin\theta\), the truck starts to roll, however firmly the brakes are set.
- Always drawing friction up the slope.Friction points up the slope only while the block tends to slide down. A pull larger than \(mg\sin\theta\) reverses both the tendency and the friction.
- Using \(\mu_s mg\) for the limit on an incline.The normal force on an incline is \(mg\cos\theta\), so the limit is \(\mu_s mg\cos\theta\).
- Forgetting that the least pull may be zero.If \(\tan\theta \le \mu_s\), the block rests by itself, and \(P_{\min}\) is zero, not negative.