Question

At what angle must the 400-lb force be applied in
order that the resultant R of the two forces have a
magnitude of 1000 lb? For this condition what will be the
angle between R and the horizontal?

Answer 1

Answer:

θ=66.42°

β=23.5781°

Explanation:

component in y: Ry = 400 Lb

Resultant = R  = 1000 Lb

to calculate the angle θ we use the cosine function

cosθ=400/100

θ=arccos(400/100)=66.42°

remember that the angles of internal of a triangle measure 180 degrees

β =180°-66.42°-90°( right angle)

β=23.5781°

Explanation:

Answer 2

Given:

R=1000N,$F_{1}$=400N,$F_{2}$=700N

α=?,β=?,θ=?

Explanation:

The Law of Cosines is used to find the remaining parts of an oblique (non-right) triangle when either the lengths of two sides and the measure of the included angle is known (SAS) or the lengths of the three sides (SSS) are known. ... The Law of Cosines states: $c^{2}$=$a^{2}$+$b^{2}$−2ab cosC .

$F_{2} ^{2}$= $R^{2}$+ $F_{1} ^{2}$– 2ab cosα

$700^{2}$=$1000^{2}$+$400^{2}$-2(1000)(400)cosα

 moves to right side to left side in negative value and cosα moves from left side divide to right side positive value.

cosα=$700^{2}$-$1000^{2}$-$400^{2}$/-2(1000)(400)

cosα=490,000-1000,000-160,000/-2(400,000)

cosα=0.8375

cos moves left side to right side into cos inverse.

α=$cos^{-1}$(0.8375)

α=33°

sinβ/400=sinα/700

400 moves to left side divide to right side in multiply value.

sinβ=(400)sinα/700

take value of α=33°,

sinβ=(400)sin33°/700

sin moves left side to right side into sin inverse.

β=$sin^{-1}$((400)sin33°/700)

β=18.13°

θ=α+β

θ=33°+18.13°

θ=51.13°

The value of α=33°, β=18.13°, θ=51.13°.