## Question 1 > I am crouching down in a corner of a wide north-south hallway in the Pentagon. A security camera is 12 m north of me, 4 m above me and 3 m west of me. How far away from me is the security camera? > [!HELP]- Solution >$ >C=(12,4,3) \qquad |C|=\sqrt{ 12^2 + 4^2+3^2}=\boxed{13m} >$ ## Question 2 > A parachutist would be descending vertically at a speed of $4ms^{-1}$ if the air were still. If a light southerly wind makes the air move horizontally at $1.5ms^{-1}$ , what will the parachutist's speed be relative to the ground? > [!HELP]- Solution >$ >A_{x}=0 \qquad A_{y}=4 \qquad W_{x}=1.5 \qquad W_{y}=0 >$ >$ >|A+W|=\sqrt{ 1.5^2 + 4^2}=\boxed{4.27ms^{-1}} >$ ## Question 3 > A mass of 50 kg is at rest on a smooth (frictionless) plane which is inclined at $20\textdegree$ to the horizontal. It is held in place by a horizontal force. Calculate the value of this force and the perpendicular reaction between the mass and the plane. > [!HELP]- Solution >$ >m=50kg \qquad \theta=20\textdegree >$ >$ >F_{x}=mg\sin\theta=\boxed{167N} \qquad R_{y}=mg\cos\theta=\boxed{461N} >$ ## Question 4 > Five forces are in equilibrium acting at a point in a horizontal plane, Four of the forces are 1 N at 0°, 2 N at 60°, 3 N at 120°, and 4 N at 210°. (These angles are measured anticlockwise from due north.) Find the magnitude and direction of the fifth force. > [!HELP]- Solution >$ >A_{x}=1\cos 0 \qquad A_{y}=1\sin 0 >$ >$ >B_{x}=2\cos 60 \qquad B_{y}=2\sin 60 >$ >$ >C_{x}=3\cos 120 \qquad C_{y}=3\sin 120 >$ >$ >D_{x}=4\cos 210 \qquad D_{y}=4\sin 210 >$ >$ >F=\sum A,B,C,D=(-2.96,\ 2.33) \qquad E=-F >$ >$ >|E|=\sqrt{ 2.96^2 + (-2.33)^2}=\boxed{3.77N} >$ >$ >\angle E=360+\tan^{-1}\left(\frac{-2.33}{2.96} \right)=\boxed{321.8\textdegree} >$ ## Question 5 > A mass m = 15 kg is supported by three cords A,B and C as in the diagram below. What are the tensions in each of the cords? > [!HELP]- Solution >$ >m=15kg \qquad T_{c}=mg=\boxed{147N} >$ >$ >T_{a}\cos 28 + T_{b}\cos 47=0 \qquad T_{a}\sin 28+T_{b}\sin 47=T_{c} >$ >$ >T_{a} =\frac{\cos47}{\cos 28}\times T_{b}=0.772\cdot T_{b} >$ >$ >0.772 \cdot T_{b}+\sin 47\cdot T_{b}=147N >$ >$ >T_{b}=\frac{147}{0.772+\sin 47}=\boxed{134.5N} >$ >$ >T_{a}=0.772\times 134.5=\boxed{103.8N} >$ ## Question 6 > Three chimpanzees are fighting over a car tyre, leading to a tug-of-war in two dimensions. The tyre is in equilibrium with Aaron exerting a force of 220 N, while Celia pulls with a force of 170 N. The direction of Celia’s pull isn’t known, but the angle between Aaron’s force and Betty’s is 137$\textdegree$. What is the magnitude of Betty’s force? > [!HELP]- Solution > Since the forces are in equilibrium, we can rearrange them to form a triangle (vectors in an equilibrium sum to zero), where the length of the sides is the magnitude, then use the cosine rule to calculate the length of the known side. > > $ > c^2=a^2+b^2-2ab\cos\theta \qquad \theta=180-137=42 > $ > $ > 170^2=220^2+F_{B}^2-2(220)(F_{B})\cos 43 > $ > $ > F_{B}^2-321.8F_{B}+19500=0 > $ > $ > F_{B}=\frac{321.8\pm \sqrt{ (-321.8)^2-4(1)(-19500) }}{2} > $ > $ > F_{B}=241N \qquad F_{B}=81N > $