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Two donkeys are tied to the same pole. One donkey pulls the pole at a strength of 5N in a direction that is a 50^(@) rotation from the east. The other donkey pulls the pole at a strength of 4N in a direction that is a 170^(@) rotation from the east. What is the combined strength of the donkeys' pulls? Round your answer to the nearest tenth. You can round intermediate values to the nearest hundredth. N

Two donkeys are tied to the same pole.\newlineOne donkey pulls the pole at a strength of 5 N 5 \mathrm{~N} in a direction that is a 50∘ 50^{\circ} rotation from the east.\newlineThe other donkey pulls the pole at a strength of 4 N 4 \mathrm{~N} in a direction that is a 170∘ 170^{\circ} rotation from the east.\newlineWhat is the combined strength of the donkeys' pulls?\newlineRound your answer to the nearest tenth. You can round intermediate values to the nearest hundredth.\newlineN \mathrm{N}

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Q. Two donkeys are tied to the same pole.\newlineOne donkey pulls the pole at a strength of 5 N 5 \mathrm{~N} in a direction that is a 50∘ 50^{\circ} rotation from the east.\newlineThe other donkey pulls the pole at a strength of 4 N 4 \mathrm{~N} in a direction that is a 170∘ 170^{\circ} rotation from the east.\newlineWhat is the combined strength of the donkeys' pulls?\newlineRound your answer to the nearest tenth. You can round intermediate values to the nearest hundredth.\newlineN \mathrm{N}
  1. Representing the forces as vectors: Represent the forces as vectors.\newlineThe first donkey pulls with a force of 5 N5\,\text{N} at a 50°50° angle from the east. This can be represented as a vector with components (5⋅cos⁡(50°),5⋅sin⁡(50°))(5 \cdot \cos(50°), 5 \cdot \sin(50°)).\newlineThe second donkey pulls with a force of 4 N4\,\text{N} at a 170°170° angle from the east. This can be represented as a vector with components (4⋅cos⁡(170°),4⋅sin⁡(170°))(4 \cdot \cos(170°), 4 \cdot \sin(170°)).
  2. Calculating the first donkey's force components: Calculate the xx and yy components of the first donkey's force.\newlineFor the first donkey:\newlinexx-component =5×cos⁡(50°)≈5×0.64279≈3.21395= 5 \times \cos(50°) \approx 5 \times 0.64279 \approx 3.21395\newlineyy-component =5×sin⁡(50°)≈5×0.76604≈3.8302= 5 \times \sin(50°) \approx 5 \times 0.76604 \approx 3.8302
  3. Calculating the second donkey's force components: Calculate the xx and yy components of the second donkey's force.\newlineFor the second donkey:\newlinexx-component =4×cos⁡(170°)≈4×(−0.98481)≈−3.93924= 4 \times \cos(170°) \approx 4 \times (-0.98481) \approx -3.93924\newlineyy-component =4×sin⁡(170°)≈4×0.17365≈0.6946= 4 \times \sin(170°) \approx 4 \times 0.17365 \approx 0.6946
  4. Finding the resultant vector: Add the xx-components and yy-components of both vectors to find the resultant vector.\newlineResultant xx-component = 3.21395+(−3.93924)≈−0.725293.21395 + (-3.93924) \approx -0.72529\newlineResultant yy-component = 3.8302+0.6946≈4.52483.8302 + 0.6946 \approx 4.5248
  5. Calculating the magnitude of the resultant vector: Calculate the magnitude of the resultant vector to find the combined strength of the donkeys' pulls.\newlineMagnitude = (x-component2+y-component2)\sqrt{(x\text{-component}^2 + y\text{-component}^2)}\newlineMagnitude ≈((−0.72529)2+(4.5248)2)\approx \sqrt{((-0.72529)^2 + (4.5248)^2)}\newlineMagnitude ≈(0.52605+20.4738)\approx \sqrt{(0.52605 + 20.4738)}\newlineMagnitude ≈(20.99985)≈4.583\approx \sqrt{(20.99985)} \approx 4.583
  6. Rounding the magnitude of the resultant vector: Round the magnitude of the resultant vector to the nearest tenth. The combined strength of the donkeys' pulls ≈4.6 N\approx 4.6\,\text{N}

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