Force — Question 6

Question 6

A bounded pair of force components gives a bounded magnitude

Independent hard bounds are $3\le F_x\le5$ N and $-4\le F_y\le-2$ N.

1. Find the smallest and largest possible magnitudes.

2. Find the possible direction angles relative to positive x.

3. Can the resultant point along an axis or vanish?

Question 6 - Solution

A bounded pair of force components gives a bounded magnitude

(a) Minimize and maximize squared components.

$$\boxed{F_{\min}=\sqrt{3^2+(-2)^2}=\sqrt{13}},\quad\boxed{F_{\max}=\sqrt{5^2+(-4)^2}=\sqrt{41}}\quad\mathrm N$$ The nearest corner to zero is $(3,-2)$ N, giving $\sqrt{13}=3.606$ N. The farthest is $(5,-4)$ N, giving $\sqrt{41}=6.403$ N.

(b) Bound the component ratio.

$$-\tan^{-1}(4/3)\le\theta\le-\tan^{-1}(2/5),\quad -53.130^\circ\le\theta\le-21.801^\circ$$ Angles lie between $-\tan^{-1}(4/3)=-53.130^\circ$ and $-\tan^{-1}(2/5)=-21.801^\circ$. Both occur at allowed corners.

(c) Use the nonzero component bounds.

$$F_x\ge3>0,\quad F_y\le-2<0\ \Rightarrow\ F\ge\sqrt{13}\,\mathrm N>0$$ Neither component can be zero, so no axis-aligned or zero force is allowed. These are guaranteed bounds, not standard deviations or confidence intervals.