Question

Find the value of F1 + F2 + F3.
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Answer 1

F = 0.78[N]

Step-by-step explanation:

The given values correspond to forces, we must remember or take into account that the forces are vector quantities, that is, they have magnitude and direction. Since we have two X-Y coordinate axes (two-dimensional), we are going to decompose each of the forces into the X & y components.

For F₁

`F_(y)=2[N]`

For F₂

`F_(x)=2*cos(60)\\F_(x)=1[N]\\F_(y)=-2*sin(60)\\F_(y)=-1.73[N]`

For F₃

`F_(x)=-1*sin(60)\\F_(x)=-0.866[N]\\F_(y)=1*cos(60)\\F_(y)=0.5 [N]`

Now we can sum each one of the forces in the given axes:

`F_(x)=1-0.866=0.134[N]\\F_(y)=2-1.73+0.5\\F_(y)=0.77[N]`

Now using the Pythagorean theorem we can find the total force.

`F=\sqrt{(0.134)^(2) +(0.77)^(2)}\\F= 0.78[N]`