The barge moves at an angle 47.3°North of West.
The tug boats exert forces in different directions on the barge and the barge moves in the direction of the resultant force.
Draw a diagram representing the forces, as shown in the figure attached. The tug boat 1 exerts a force F₁ along North while the tug boat 2 exerts a force F₂ at an angle of 15° North of west.
Resolve the forces along the North and the West.
The force acting along the North is given by,
[tex]F_N=F_1+F_2sin15\\ =(7.5*10^4N)+(9.5*10^4 N*sin15)\\ =9.96*10^4N[/tex]
The force along the West is given by,
[tex]F_W=F_2cos15=(9.5*10^4N)cos15\\ =9.18*10^4N[/tex]
The resultant force F makes an angle θ North of West.
Therefore, from the diagram,
[tex]tan\theta =\frac{F_N}{F_W} \\ =\frac{9.96*10^4N}{9.18*10^4N} \\ =1.0849[/tex]
Hence,
[tex]\theta =tan^-^1(1.00849)\\ =47.3^o[/tex]
Thus, the barge moves in a direction 47.3° North of West.
The barge move due to the pulling force of two boats as 47.3 degrees in north-west direction.
Force is the effect of pull or push due to which the object having a mass changes its velocity.
The force is of two types-
Two tugboats pull a barge across the harbor.
In the given situation, one boat exerts a force of [tex]7.5 \times 10^4\rm N[/tex]in the direction of north and the the second boat exerts a force of [tex]9.5 \times 10^4\rm N[/tex]N at 15 degrees north in the direction of west.
For this situation the total force acting on the north direction can be given as,
[tex]F_N=7.5\times10^4+9.5\times10^4(\sin 15^o)\\F_N=7.5\times10^4+2.459\times10^4\\F_N=9.96\times10^4[/tex]
Now the total force acting on the west direction can be given as,
[tex]F_W=9.5\times10^4(\cos15^o)\\F_W=9.18\times10^4[/tex]
The tangent theta for this situation can be given as,
[tex]\tan \theta=\dfrac{9.96\times10^{4}}{9.18\times10^{4}}\\\theta=\tan^{-1}(1.0849)\\\theta=47.3^o[/tex]
Thus, the barge move due to the pulling force of two boats as 47.3 degrees in north-west direction.
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