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A basket of flowers weighing $2$kg is placed on a flat grassy slope. The coefficient of static friction between the basket and the slope is $0.4$, and the basket is on the point of slipping down the slope.

Write down vector expressions for the forces and the equilibrium condition. Hence determine the angle that the slope makes with the horizontal.

So I've started with drawing a force diagram (including friction), where $F$ is acting in the $i$ direction, so $F=Fi$, $W$ is the weight $=2$kg and $N$ is the normal reaction.

How do I determine the angle?

levap
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S17
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2 Answers2

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enter image description here $\small{\text{(the downslope force is a component of gravity}}$ $\small{\text{acting on mass, not a separate force,}}$ $\small{\text{shown for later use in the formula.}}$ $\small{\text{Force values are magnitudes so unsigned).}}$

To start sliding, you need the force due to mass downslope, $mg \sin \theta$ to exceed the resistive friction, $F_f = \mu N = \mu mg \cos\theta$. At the threshold slope the two values are equal. So here we need

$$2g \sin \theta = 0.4 \cdot 2g \cos\theta \\ \implies \tan\theta = 0.4$$

So $\theta = \tan^{-1} 0.4 \approx 21.8°$. I'm ignoring static vs kinetic friction, because you've only been given one value anyway.

Joffan
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Define $\theta$ as the angle of the slope relative to horizontal.

You should get a Normal force of $mg\cos\theta$ so that the force of Friction will be $(0.4)mg\cos\theta$ ( acting up the slope )

The component of Gravity acting down the slope will be $mg\sin\theta$

Solve for equilibrium $\theta$ by making these two forces balance ...

$$ (0.4)mg\cos\theta = mg\sin\theta $$

WW1
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