Physics, 29.06.2019 01:30 jayzelgaspar8441

Two tugboats pull a barge across the harbor. one boat exerts a force of 7.5 x 10^4n north, while the second boat exerts a force of 9.5 x 10^4n at 15 degrees north of west. precisely, in what direction does the barge move?

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Answer from: CoreyHammond584

The barge move due to the pulling force of two boats as 47.3 degrees in north-west direction.

What is force?

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-

Push-When the force applied in the direction of motion of the object, then the force is called the push force. Pull- When the force applied in the opposite direction of motion of the object, then the force is called the pull force.

Two tugboats pull a barge across the harbor.

In the given situation, one boat exerts a force of $7.5 \times 10^4\rm N$ in the direction of north and the the second boat exerts a force of $9.5 \times 10^4\rm N$ 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,

$F_N=7.5\times10^4+9.5\times10^4(\sin 15^o)\\F_N=7.5\times10^4+2.459\times10^4\\F_N=9.96\times10^4$

Now the total force acting on the west direction can be given as,

$F_W=9.5\times10^4(\cos15^o)\\F_W=9.18\times10^4$

The tangent theta for this situation can be given as,

$\tan \theta=\dfrac{9.96\times10^{4}}{9.18\times10^{4}}\\\theta=\tan^{-1}(1.0849)\\\theta=47.3^o$

Thus, the barge move due to the pulling force of two boats as 47.3 degrees in north-west direction.

Learn more about the force here;

link

Two tugboats pull a barge across the harbor. one boat exerts a force of 7.5 x 10^4n north, while the

Answer from: tristan4233

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,

$F_N=F_1+F_2sin15\\ =(7.5*10^4N)+(9.5*10^4 N*sin15)\\ =9.96*10^4N$

The force along the West is given by,

$F_W=F_2cos15=(9.5*10^4N)cos15\\ =9.18*10^4N$

The resultant force F makes an angle θ North of West.

Therefore, from the diagram,

$tan\theta =\frac{F_N}{F_W} \\ =\frac{9.96*10^4N}{9.18*10^4N} \\ =1.0849$

Hence,

$\theta =tan^-^1(1.00849)\\ =47.3^o$

Thus, the barge moves in a direction 47.3° North of West.

Answer from: Quest

massxgravity x heght=weightxheight= pe

4900=490xh

h=10

Answer from: Quest

Iwannasay c is the most accurate

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