Mathematics 9709/42 — October/November 2010

Walkthrough

The block is on horizontal ground, so the normal reaction $R$ acts perpendicular to the ground. The applied force of magnitude $2000\,\text{N}$ is at $15^{\circ}$ to the upwards vertical, so its vertical component is $2000\cos 15^{\circ}$ upward and its horizontal component is $2000\sin 15^{\circ}$.

Because the block is in limiting equilibrium, the resultant force in every direction is zero, and friction has its maximum possible value. This means we can use $F = \mu R$, not just $F \leq \mu R$.

First resolve vertically. The upward forces are the vertical component $2000\cos 15^{\circ}$ and the normal reaction $R$. Together they balance the weight $400g$. This gives $R + 2000\cos 15^{\circ} = 400g$, so the normal reaction is smaller than the weight because the applied force helps support the block.

Next resolve horizontally. The only horizontal forces are the horizontal component of the applied force and the friction. Since the block is in equilibrium, $F = 2000\sin 15^{\circ}$.

Finally substitute into $F = \mu R$ and solve for $\mu$. The units cancel, giving a dimensionless coefficient.

Key Takeaways

• Forces must be resolved into perpendicular components before applying equilibrium.
• When an angle is given to the vertical, the component along the vertical uses cosine and the component perpendicular to it uses sine.
• In limiting equilibrium, friction is at its maximum value and satisfies $F = \mu R$.
• The normal reaction is not always equal to the weight; it can be reduced by an upward applied force.

Common Mistakes

• Interchanging sine and cosine: using $2000\sin 15^{\circ}$ for the vertical component and $2000\cos 15^{\circ}$ for the horizontal component gives the incorrect coefficient $0.55$ and would lose a mark.
• Treating the force as $15^{\circ}$ above the horizontal instead of $15^{\circ}$ to the vertical.
• Forgetting that the upward component reduces the normal reaction, so writing $R = 400g$ without the component is incorrect.
• Using $F \leq \mu R$ as an inequality instead of the equality that applies in limiting equilibrium.

Things to Be Careful About

• Use the value of $g$ stated in the question. In this mark scheme, taking $g = 10\,\text{m s}^{-2}$ gives $\mu = 0.25$; using $g = 9.8\,\text{m s}^{-2}$ would give a different value.
• The final answer must be given correct to 2 significant figures, so write $0.25$, not $0.2503$ or $0.250$.
• The coefficient of friction is dimensionless; do not attach units such as N or kg.
• When substituting into $F = \mu R$, use the normal reaction $R$ found from the vertical resolution, not the weight of the block.