1) X-component: 568.5 N, Y-component: 511.9 N
2) The horizontal force is 86.6 N and the resulting acceleration is [tex]0.87 m/s^2[/tex]
Explanation:
1)
In this first part of the problem, we have to resolve the force into its two components, along the x and the y direction.
The two components are given by:
[tex]F_x = F cos \theta\\F_y = F sin \theta[/tex]
where
F = 765 N is the magnitude of the force
[tex]\theta=42.0^{\circ}[/tex] is the angle of the force with the horizontal
Substituting, we find:
- Horizontal component: [tex]F_x = (765)(cos 42.0^{\circ})=568.5 N[/tex]
- Vertical component: [tex]F_y = (765)(sin 42.0^{\circ})=511.9 N[/tex]
2)
First of all, we have to find the horizontal component of the pulling force, which is given by
[tex]F_x = F cos \theta[/tex]
where
F = 100 N is the magnitude of the pulling force
[tex]\theta=30.0^{\circ}[/tex] is the direction of the force with the horizontal
Substituting,
[tex]F_x = (100)(cos 30^{\circ})=86.6 N[/tex]
Now we can find the acceleration of the wagon by using Newton's second law:
[tex]F_x = ma_x[/tex]
where
[tex]F_x = 86.6 N[/tex] is the net force in the horizontal direction
m = 100 kg is the mass of the wagon
[tex]a_x[/tex] is the acceleration in the horizontal direction
Solving for [tex]a_x[/tex], we find
[tex]a_x = \frac{F_x}{m}=\frac{86.6}{100}=0.87 m/s^2[/tex]
Learn more about vector components:
brainly.com/question/2678571
And about Newton's second law:
brainly.com/question/3820012
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