Paper 3 Forces Answers

Q1 Equilibrium of a signboard

1. The hinge force has unknown components but its line of action passes through the hinge. Taking moments about the hinge removes the hinge force from the moment equation. [2]
2. Only the vertical component of cable tension produces a moment about the hinge:

$$T\sin 35^{\circ }(1.50)=90.0(1.10).$$

Hence

$$T=\frac{99.0}{1.50\sin 35^{\circ }}=115\mathrm{N}.$$

[4] 3. Horizontal equilibrium: hinge horizontal component $=T\cos 35^{\circ }=94.1\mathrm{N}$, opposite to the cable’s horizontal pull. Vertical equilibrium: $H_y+T\sin 35^{\circ }-90.0=0$. Since $T\sin 35^{\circ }=66.0\mathrm{N}$, $H_y=24.0\mathrm{N}$ upward. [4] 4. Examples: board is light; cable and hinge are ideal; board is rigid; the hanging sign’s weight acts at a point; static equilibrium holds. [1]

Q2 Object falling through a fluid

1. Initial drag is zero. Resultant force downward $=0.420-0.060=0.360\mathrm{N}$. [2]
2. As speed increases, resistive force increases upward. The upward forces therefore increase, reducing the downward resultant force. Since $a=F_{\mathrm{result}}/m$, acceleration decreases. [3]
3. At terminal velocity, resultant force is zero:

$$F_{\mathrm{drag}}+0.060=0.420,$$

so $F_{\mathrm{drag}}=0.360\mathrm{N}$. [2] 4. Sketch starts at zero speed, rises with initially large gradient, then flattens towards a horizontal terminal-speed asymptote. Explanation: gradient is acceleration; acceleration decreases to zero as drag increases to balance weight minus upthrust. [4] 5. Greater viscosity gives a larger resistive force at the same speed. The balance condition is reached at a lower speed, so terminal velocity is smaller. [2]