Answer :
Answer:
a) [tex]F=429.261\ N[/tex]
b) [tex]\beta=45.8494^{\circ}[/tex] with respect to the west direction.
c) [tex]F=429.261\ N[/tex]
d) [tex]\beta'=135.85^{\circ}[/tex] with respect to the west direction.
Explanation:
Given:
- force of push on the crate in the west direction, [tex]F_w=299\ N[/tex]
- force of push on the crate in the north direction, [tex]F_n=308\ N[/tex]
- Since both the directions North and West are mutually perpendicular to each other.
a)
The magnitude of resultant force:
[tex]F=\sqrt{(299)^2+(308)^2}[/tex]
[tex]F=429.261\ N[/tex]
b)
Now the direction of the force:
[tex]\tan\beta=\frac{F_n}{F_w}[/tex]
[tex]\tan\beta=\frac{299}{308}[/tex]
[tex]\beta=45.8494^{\circ}[/tex] with respect to the west direction.
- When the second worker will apply the same magnitude of force but negative then the direction of the force will be reversed.
c) The magnitude of the resultant will remain the same.
[tex]F=\sqrt{(299)^2+(-308)^2}[/tex]
[tex]F=429.261\ N[/tex]
d)
For the angle with respect to the west:
[tex]\tan\phi=\frac{308}{299}[/tex]
[tex]\phi=45.84^{\circ}[/tex]
Now from the schematic:
[tex]\beta'=90+45.85[/tex]
[tex]\beta'=135.85^{\circ}[/tex] with respect to the west direction.