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Given forces: F1 = 7 N and F2 = 10 N.

Mathematics

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4 parts
a

17 N\text{17 N}

b

3 N\text{3 N}

c

14.80 N\text{14.80 N}

d

4.18 N\text{4.18 N}

Step-by-step solution

8 steps

The resultant of two vectors is a single vector that produces the same effect as the original two vectors combined. It is the vector sum of the two vectors.

Given forces: F1=7F_1 = 7 N and F2=10F_2 = 10 N.
The magnitude of the resultant force RR of two forces F1F_1 and F2F_2 with an angle θ\theta between them is given by the formula:

R=F12+F22+2F1F2cos⁡θR = \sqrt{F_1^2 + F_2^2 + 2F_1F_2 \cos\theta}

Part a· 2 steps

The forces are parallel and act in the same direction.

  1. Determine the angle between the forces.

    When forces are parallel and act in the same direction, θ=0∘\theta = 0^\circ.

  2. Substitute the values into the formula.

    R=(7 N)2+(10 N)2+2(7 N)(10 N)cos⁡(0∘)=49 N2+100 N2+140 N2(1)=289 N2=17 N\begin{align*} R &= \sqrt{(7 \text{ N})^2 + (10 \text{ N})^2 + 2(7 \text{ N})(10 \text{ N}) \cos(0^\circ)} \\ &= \sqrt{49 \text{ N}^2 + 100 \text{ N}^2 + 140 \text{ N}^2 (1)} \\ &= \sqrt{289 \text{ N}^2} \\ &= 17 \text{ N}\end{align*}

    Alternatively, for forces in the same direction, R=F1+F2R = F_1 + F_2.

    R=7 N+10 N=17 NR = 7 \text{ N} + 10 \text{ N} = 17 \text{ N}

    The resultant force is 17 N\boxed{\text{17 N}}.

Part b

3 N\text{3 N}

The forces are parallel and act in opposite directions.

  1. Determine the angle between the forces.

    When forces are parallel and act in opposite directions, θ=180∘\theta = 180^\circ.

  2. Substitute the values into the formula.

    R=(7 N)2+(10 N)2+2(7 N)(10 N)cos⁡(180∘)=49 N2+100 N2+140 N2(−1)=149 N2−140 N2=9 N2=3 N\begin{align*} R &= \sqrt{(7 \text{ N})^2 + (10 \text{ N})^2 + 2(7 \text{ N})(10 \text{ N}) \cos(180^\circ)} \\ &= \sqrt{49 \text{ N}^2 + 100 \text{ N}^2 + 140 \text{ N}^2 (-1)} \\ &= \sqrt{149 \text{ N}^2 - 140 \text{ N}^2} \\ &= \sqrt{9 \text{ N}^2} \\ &= 3 \text{ N}\end{align*}

    Alternatively, for forces in opposite directions, R=∣F2−F1∣R = |F_2 - F_1|.

    R=∣10 N−7 N∣=3 NR = |10 \text{ N} - 7 \text{ N}| = 3 \text{ N}

    The resultant force is 3 N\boxed{\text{3 N}}.

Part c

14.80 N\text{14.80 N}

The two forces are inclined at an angle of 60∘60^\circ to each other.

  1. Determine the angle between the forces.

    θ=60∘\theta = 60^\circ.

  2. Substitute the values into the formula.

    R=(7 N)2+(10 N)2+2(7 N)(10 N)cos⁡(60∘)=49 N2+100 N2+140 N2(0.5)=149 N2+70 N2=219 N2≈14.80 N\begin{align*} R &= \sqrt{(7 \text{ N})^2 + (10 \text{ N})^2 + 2(7 \text{ N})(10 \text{ N}) \cos(60^\circ)} \\ &= \sqrt{49 \text{ N}^2 + 100 \text{ N}^2 + 140 \text{ N}^2 (0.5)} \\ &= \sqrt{149 \text{ N}^2 + 70 \text{ N}^2} \\ &= \sqrt{219 \text{ N}^2} \\ &\approx 14.80 \text{ N}\end{align*}

    The resultant force is 14.80 N\boxed{\text{14.80 N}}.

Part d

4.18 N\text{4.18 N}

The two forces are inclined at an angle of 160∘160^\circ.

  1. Determine the angle between the forces.

    θ=160∘\theta = 160^\circ.

  2. Substitute the values into the formula.

    R=(7 N)2+(10 N)2+2(7 N)(10 N)cos⁡(160∘)=49 N2+100 N2+140 N2(−0.9397)=149 N2−131.558 N2=17.442 N2≈4.18 N\begin{align*} R &= \sqrt{(7 \text{ N})^2 + (10 \text{ N})^2 + 2(7 \text{ N})(10 \text{ N}) \cos(160^\circ)} \\ &= \sqrt{49 \text{ N}^2 + 100 \text{ N}^2 + 140 \text{ N}^2 (-0.9397)} \\ &= \sqrt{149 \text{ N}^2 - 131.558 \text{ N}^2} \\ &= \sqrt{17.442 \text{ N}^2} \\ &\approx 4.18 \text{ N}\end{align*}

    The resultant force is 4.18 N\boxed{\text{4.18 N}}.

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