Question

A rocket fires two engines simultaneously. One produces a thrust of 685 N directly forward while the other gives a thrust of 507 N at an angle 33.4 ∘ above the forward direction.Find the magnitude of the resultant force which these engines exert on the rocket.Find the deviation of the direction (relative to the forward direction) of the resultant force which these engines exert on the rocket.

Answer 1

Sol:

Given thrust of 685 N directly forward
other thrust of 507 N at an angle 33.4 ∘ above

Let

X axis = forward direction

resultant force exert on the rocket
R = F₁ + F₂
Now
X component of F₁ will be

Fx₁ = 685 N
X component of F₂
Fx₂ = 507 * cos 33.4°
= 423.267 N  
Y component of F₁
Fy₁ = 0 N
Y component of F₂
Fy₂ = 507* sin 33.4°
= 279.093 N

Rx = Fx₁ + Fx₂
= 685 N + 423.267 N
= 1 108.267 N

resultant X component
Rx = 1 108.267 N

Ry = Fy₁ + Fy₂
= 0 N + 273.805 N
= 273.805 N

resultant Y component

Ry = 279.093 N

R = √[(Rx₁)² + (Ry₂)²]
= √[(1 108.267 )² + (279.093 )²]
= 1 142.862 N

resultant force exert on the rocket
R = 1 142.862 N

θ = atan (Ry / Rx)
= atan (279.093/ 1 108.267)
= 14.13°

resultant force deviation angle
θ = 14.13°

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