Answer:
F₄ = 29.819 N
Explanation:
Given
F₁ = (- 25*Cos 50° i + 25*Sin 50° j + 0 k) N
F₂ = (12*Cos 50° i + 12*Sin 50° j + 0 k) N
F₃ = (0 i + 0 j + 4 k) N
Then we have
F₁ + F₂ + F₃ + F₄ = 0
⇒ F₄ = - (F₁ + F₂ + F₃)
⇒ F₄ = - ((- 25*Cos 50° i + 25*Sin 50° j) N + (12*Cos 50° i + 12*Sin 50° j) N + (4 k) N) = (13*Cos 50° i - 37*Sin 50° j - 4 k) N
The magnitude of the force will be
F₄ = √((13*Cos 50°)² + (- 37*Sin 50°)² + (- 4)²) N = 29.819 N
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Answer:
![d = 3.19 km](https://tex.z-dn.net/?f=d%20%3D%203.19%20km)
direction is given as
South of West
Explanation:
Part b)
displacement is given as
![d_1 = 3.1 \hat j](https://tex.z-dn.net/?f=d_1%20%3D%203.1%20%5Chat%20j)
![d_2 = 2.4 \hat i](https://tex.z-dn.net/?f=d_2%20%3D%202.4%20%5Chat%20i)
![d_3 = 5.2(-\hat j)](https://tex.z-dn.net/?f=%20d_3%20%3D%205.2%28-%5Chat%20j%29)
now we will have
![d = d_1 + d_2 + d_3](https://tex.z-dn.net/?f=d%20%3D%20d_1%20%2B%20d_2%20%2B%20d_3)
![d = 2.4 \hat i + (3.1 - 5.2)\hat j](https://tex.z-dn.net/?f=d%20%3D%202.4%20%5Chat%20i%20%2B%20%283.1%20-%205.2%29%5Chat%20j)
![d = 2.4 \hat i - 2.1 \hat j](https://tex.z-dn.net/?f=d%20%3D%202.4%20%5Chat%20i%20-%202.1%20%5Chat%20j)
total displacement is given as
![d = \sqrt{2.4^2 + 2.1^2}](https://tex.z-dn.net/?f=d%20%3D%20%5Csqrt%7B2.4%5E2%20%2B%202.1%5E2%7D)
![d = 3.19 km](https://tex.z-dn.net/?f=d%20%3D%203.19%20km)
direction is given as
![tan\theta = \frac{-2.1}{2.4}](https://tex.z-dn.net/?f=tan%5Ctheta%20%3D%20%5Cfrac%7B-2.1%7D%7B2.4%7D)
South of West