Q3.1 Forces are vectors (S)

A body of mass 2 kg experiences three forces, as shown. Forces $\vec{F}_1$ and $\vec{F}_2$ each have magnitude 10 N.
  1. Determine the value of F3 if the system is in static equilibrium.
  2. Determine the acceleration of the mass if F3=10 N.
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Solution

  1. Saying that the system is in static equilibrium means the body is is (and remains) stationary: thus it has zero acceleration and by Newton’s 2nd law (Key Point 2.2) must experience zero net force. Let x and y denote horizontal and vertical directions. Then (since forces are vectors!) both the x and y components of the net force must vanish.

    By inspection the y components of the forces $\vec{F}_1$ and $\vec{F}_2$ are equal and opposite; and $\vec{F}_3$ has no vertical component. So there is indeed no net y-component.

    To ensure that there is no x-component we require that $\vec{F}_3$ (entirely in the x-direction) is balanced by the sum of the x-components of $\vec{F}_1$ and $\vec{F}_2$. Thus

    \[ F_3= F_1 \cos{30} + F_2 \cos{30} = 2 \times 10 \times \frac{\sqrt {3}}{2} = 10\sqrt{3} \, N \]

  2. In this case the horizontal components of the forces no longer balance. There will be a net force in the x direction and thus an acceleration (also in the x direction) given by
    \[ ma = F_3 - F_1 \cos{30} - F_2 \cos{30}= 10 (1-\sqrt{3}) \]
    implying
    \[ a = -3.66 \,ms^{-2}\hspace{0.5cm}\mbox{\rm (3sf)}\hspace{0.5cm} \]
    The minus sign means that the acceleration is to the left.