1)
From the information given, the magnitude of the forces and their direction are
A(2N, 60degrees)
B(4.5N, 180degrees)
C(5N, 30degrees)
We would resolve the vectors into x and y components. Recall,
x component = FCosθi
y component = FSinθj
where
F is the force
θ is the angle
Considering A(2N, 60degrees),
F = 2, θ = 60.
x component = 2cos60i = 2 x 0.5i = i
y component = 2sin60i = 1.7321j
Considering B(4.5N, 180degrees),
F = 4.5, θ = 180.
x component = 4.5cos180i = - 4.5i
y component = 4.5sin180 = 0j
Considering C(5N, 300degrees),
F = 5, θ = 300
x component = 5cos300i = 2.5i
y component = 5sin300 = - 4.3301j
R = Σi + Σj
where
R is the resultant force
Σi = i - 4.5i + 2.5i = - i
Σj = 1.7321j + 0j - 4.3301j = - 2.5981j
magnitude of the resultant force is |R|
|R| = √(Σi)^2 + (Σj)^2 = √(- 1)^2 + (- 2.5981)^2
|R| = 2.78 N
The resultant force of the body is 2.78N
2) Recall the formula for calculating force;
F = ma
where
m is the mass of the body
F is the force
a is the acceleration
From the information given,
m = 2
F = 2.78
Thus,
2.78 = 2a
a = 2.78/2
a = 1.39 m/s^2
The acceleration with which the body begins to move is 1.39 m/s^2
