The figure is missing: find it in attachment.
a) (3i - 5j) N
First of all, we have to write each force as a vector with a direction:
- Force F1 points downward, so along the negative y-direction, so we can write it as
[tex]F_1 = -5 j[/tex]
- Force F2 points to the right, so along the positive x-direction, so we can write it as
[tex]F_2 = +8 i[/tex]
- Force F3 points to the left, so along the negative x-direction, so we can write it as
[tex]F_3 = -5 i[/tex]
Now we can write the net force, by adding the three vectors and separating the x-component from the y-component:
[tex]F=F_1+F_2+F_3 = -5j+8i-5i = (8-5)i+(-5)j=(3i-5j)N[/tex]
b) 5.8 N
The magnitude of the net force can be calculated by applying Pythagorean's theorem on the components of the net force:
[tex]F=\sqrt{F_x^2+F_y^2}[/tex]
where
[tex]F_x = 3 N[/tex] is the x-component
[tex]F_y = -5 N[/tex] is the y-component
Substituting into the equation,
[tex]F=\sqrt{(3)^2+(-5)^2}=5.8 N[/tex]
c) [tex]-59.0^{\circ}[/tex]
The angle of the net force, measured with respect to the positive x-axis (counterclockwise), can be calculated by using the formula
[tex]\theta=tan^{-1} (\frac{F_y}{F_x})[/tex]
where
[tex]F_x = 3 N[/tex] is the x-component
[tex]F_y = -5 N[/tex] is the y-component
Substituting into the equation, we find
[tex]\theta = tan^{-1} (\frac{-5}{3})=-59.0^{\circ}[/tex]
d) [tex]0.84 m/s^2[/tex]
The acceleration can be found by using Newton's second law:
[tex]F=ma[/tex]
where
F is the net force on an object
m is its mass
a is the acceleration
For the object in the problem, we know
F = 5.8 N
m = 6.9 kg
Solving the equation for a, we find the magnitude of the acceleration:
[tex]a=\frac{F}{m}=\frac{5.8}{6.9}=0.84 m/s^2[/tex]
A) The net force in component form is;
F_net = 3i^ - 5j^
B) The magnitude of the net force is;
F_net = 5.83 N
C) The angle of the net force is; θ = -59.04°
D) The magnitude of the acceleration is; a = 0.845 m/s²
The image of the object and the 3 forces is missing and so i have attached it.
We are given;
F₁ = 5 N
F₂ = 8 N
F₃ = 5 N
A) From the attached diagram, we see that the 3 forces are vectors where;
F₁ is vector in negative y-direction
F₂ is vector in positive x-direction
F₃ is vector in negative x-direction
Thus, we will have;
F₁ = -5j^ N
F₂ = 8i^ N
F₃ = -5i^ N
Thus;
Net force (F_net) = F₁ + F₂ + F₃
F_net = -5j^ + 8i^ + -5i^
F_net = 3i^ - 5j^
B) The magnitude of the net force is;
F_net = √((F_x)² + (F_y)²)
F_net = √((3²) + (-5)²)
F_net = √34
F_net = 5.83 N
C) Angle in degrees of the net force measured from the x-axis is;
θ = tan⁻¹(F_y/F_x)
θ = tan⁻¹(-5/3)
θ = -59.04°
D) Since we are told that block has a mass(m) of 6.9 kg, then acceleration of block is gotten from;
a = F_net/m
a = 5.83/6.9
a = 0.845 m/s²
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