# Can anyone explain this problem with an aid of a diagram?

Jan 27, 2017

The resultant force is 86.74 N at an angle of 72.1°

Sorry, no diagram!

#### Explanation:

First, you will resolve each vector (given here in standard form) into rectangular components ($x$ and $y$).

Then, you will add together the $x -$components and add together the $y -$components. This will given you the answer you seek, but in rectangular form.

Finally, convert the resultant into standard form.

Here's how:

Resolve into rectangular components (note the changes of angle measure into standard angles):

Fx_1 = 35 cos 110° = 35 (-0.342) = -11.97 N

Fy_1 = 35 sin 110° = 35 (0.940) = 32.89 N

Fx_2 = 60 cos 90° = 0 N

Fx_1 = 60 sin 90° = 60 N

Fx_3 = 40 cos -15° = 40 (0.969) = 38.64 N

Fy_3 = 40 sin -15° = 40 (-0.259) = -10.35 N

Now, add the one-dimensional components

${F}_{x} = {F}_{x 1} + {F}_{x 2} + {F}_{x 3} = 26.67 N$

and

${F}_{y} = {F}_{y 1} + {F}_{y 2} + {F}_{y 3} = 82.54 N$

This is the resultant force in rectangular form. With a positive $x$-component and a positive $y$-component, this vector points into the 1st quadrant.

Now, convert to standard form:

$F = \sqrt{{\left({F}_{x}\right)}^{2} + {\left({F}_{y}\right)}^{2}} = \sqrt{{\left(26.67\right)}^{2} + {82.54}^{2}} = 86.74 N$

theta=tan^(-1)(82.54/(26.67)) = 72.1°

Jan 27, 2017

$\text{Problem was solved again.}$

#### Explanation:

$\text{all forces acting on object}$

$\text{..................................................................................................}$

$\text{The Force "F_1" and its components(vertical,horizontal)}$

F_("1x")=-F_1.sin(20)=-35.0,34202014=-11.97" N"

F_("1y")=F_1.cos(20)=35.0,93969262=32.89" N"

$\text{....................................................................................}$

$\text{The Force "F_2" and its components(vertical,horizontal)}$

$\text{The Force "F_2 " has vertical component only .}$

${F}_{\text{2x}} = 0$

F_("2y")=60" N"

$\text{....................................................................................................}$

$\text{The Force "F_3" and its components(vertical,horizontal)}$

F_("3x")=F_3.sin(75)=40.0,96592583=38.64" N "

F_("3y")=-F_3.cos(75)=-40.0,25881905=-10.35" N"

$\ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots . .$

$\text{now let us find the total "F_x" and "F_ y " components.}$

${F}_{x} = {F}_{\text{1x")+F_("2x")+F_("3x}}$

${F}_{y} = {F}_{\text{1y")+F_("2y")+F_("3y}}$

${F}_{x} = - 11.97 + 0 + 38.64 = 26.67 \text{ N}$

${F}_{y} = 32.89 + 60 - 10.35 = 82.54 \text{ N}$

$\text{..................................................................................}$

$\text{Resultant Vector...}$

$\text{ magnitude of the resultant vector can be calculated :}$

$F = \sqrt{{\left({F}_{x}\right)}^{2} + {\left({F}_{y}\right)}^{2}}$

$F = \sqrt{{\left(26.67\right)}^{2} + {\left(82.54\right)}^{2}}$

$F = \sqrt{711.2889 + 6812.8516}$

$F = \sqrt{7524.1405}$

$F = 86.74 \text{ N}$

$\text{we must find " tan(theta) " for direction of the resultant vector.}$

$\tan \left(\theta\right) = \frac{{F}_{x}}{{F}_{y}}$

$\tan \left(\theta\right) = \frac{26.67}{82.54}$

$\tan \left(\theta\right) = 0.32311606$

$\theta = 17.91$