The angle of the force is [tex]51.1^{\circ}[/tex]
Explanation:
To solve this problem, we can apply Newton's second law along the horizontal direction of motion of the suitcase:
[tex]\sum F_x = ma_x[/tex]
where
[tex]\sum F_x[/tex] is the net force along the x-axis
m is the mass of the suitcase
[tex]a_x[/tex] is the acceleration along the x-axis
The suitcase is moving at constant speed, so the acceleration is zero:
[tex]a_x=0[/tex]
Therefore the net force must also be zero:
[tex]\sum F_x = 0[/tex] (1)
We have two forces acting along the horizontal direction:
- The component of the push (forward) in the horizontal direction, [tex]F cos \theta[/tex], with
F = 43 N
[tex]\theta[/tex] = angle of the force with the horizontal
- The force of friction, [tex]F_f = 27 N[/tex], backward
So the net force can be written as
[tex]\sum F_x = F cos \theta - F_f[/tex] (2)
Combining (1) and (2),
[tex]F cos \theta - F_f = 0[/tex]
And so we can find the angle:
[tex]\theta = cos^{-1}(\frac{F_f}{F})=cos^{-1}(\frac{27}{43})=51.1^{\circ}[/tex]
Learn more about Newton's second law:
brainly.com/question/3820012
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