Answer to Question #124453 in Mechanics | Relativity for Chisaka

Question

Find the resultant R of the following forces all acting on the same point in the given directions: 30lb northeast; 70 lb to the south; and 50 lb 20o north of west.

Answer 1

To solve task lets find projections of forces on axis North-South and West-East.

Projection of force 30lb on axis N-S:

$F_{30ns}= 30\times cos 45\degree = 30 \times 0.707 = 21.21$

this force directed to North

Projection of force 30lb on axis W-E is the same:

$F_{30we}= 30\times cos 45\degree = 30 \times 0.707 = 21.21$

this force directed to East.

Projection of force 50lb on axis W-E:

$F_{50we}=50\times cos20\degree = 50 \times 0.94 = 47$

this force directed to West

Projection of force 50lb on axis N-S:

$F_{50ns}=50\times cos70\degree = 50\times0.342=17.1$

this force directed to North.

Force 70lb completely lays on axis N-S.

Now, lets find sum of forces on axis W-E and N-E.

For W-E:

$F_{we} = F_{50we}-F_{30we}=47-21.21=25.79$

this force directed to West.

For N-S:

$F_{ns}=F_{70} - F_{30ns}-F_{50ns}=70-21.21-17.1=31.69$

this force directed to South.

Now we need to find sum of two last forces.

Because they are perpendicular to each other. We can do the following:

$F_{result}=\sqrt(F_{we}^2+F_{ns}^2)=\sqrt(25.79^2+31.69^2)=40.86$

Now lets find angle of resulting force to axis N-S:

$\alpha=arctan(\frac{F_{we}}{F_{ns}})=arctan(\frac{25.79}{31.69})=39,14\degree$