Question

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

Answer 1

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_(x1)+F_(x2) +F_(x3) = 26.67 N`

and

`F_y = F_(y1)+F_(y2) +F_(y3) = 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((F_x)^2+(F_y)^2) = sqrt((26.67)^2+82.54^2) = 86.74 N`

`theta=tan^(-1)(82.54/(26.67)) = 72.1°`

Answer 2

`"Problem was solved again."`

Explanation:

`"all forces acting on object"`

`".................................................................................................."`

`"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"`

`"...................................................................................."`

`"The Force "F_2" and its components(vertical,horizontal)"`

`"The Force "F_2 " has vertical component only ."`

`F_("2x")=0`

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

`"...................................................................................................."`

`"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"`

`.....................................................................................................`

`"now let us find the total "F_x" and "F_ y " components."`

`F_x=F_("1x")+F_("2x")+F_("3x")`

`F_y=F_("1y")+F_("2y")+F_("3y")`

`F_x=-11.97+0+38.64=26.67" N"`

`F_y=32.89+60-10.35=82.54" N"`

`".................................................................................."`

`"Resultant Vector..."`

`" magnitude of the resultant vector can be calculated :"`

`F=sqrt((F_x)^2+(F_y)^2)`

`F=sqrt((26.67)^2+(82.54)^2)`

`F=sqrt(711.2889+6812.8516)`

`F=sqrt(7524.1405)`

`F=86.74" N"`

`"we must find " tan(theta) " for direction of the resultant vector."`

`tan(theta)=(F_x)/(F_y)`

`tan(theta)=(26.67)/(82.54)`

`tan(theta)=0.32311606`

`theta=17.91`