Physics

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Two tugboats are pulling a freighter. Find the net force and direction of the acceleration for the following cases. Assume that each tugboat pulls with a force of 1.2 x 10^5 N. Assume no resistive force at this time.
a) Tugboat 1 pulls in a direction given by [N 60 degrees E] and tugboat 2 pulls in a direction given by [S 60 degrees E].

The answer says: Fnet=(1.2 x 10^5 cos 30 degrees ) x2
=2.1 x 10^5 N [E]

but I don't know know they got that.
I used the cosine law with contained angle 60 degrees, but my answer was different from the previous one.

Thanks in advance!

  • Physics -

    look at the angles, each is pulling to the side of the forward direction by 30 deg.

    The component of each rope to the side is equal and opposite, so they add to zero side force.

    Now the forward force on each rope is Tension*cos30. There are two of those.

    2*Tension*cos30deg is the net force, and of course it is in the East direction. acceleration is this force divided by the mass of the tug (neglecting friction)>

  • Physics -

    How come we don't use the cosine law? In an example, they used the cosine law to find Fnet:

    A person of mass 70 kg is sitting on a 20 kg toboggan. If the two people are pulling her with two different ropes, find the person's acceleration. The force of person 1 is 50 N [ E 40 degrees N], and the force of person 2 is 60 N [E 25 degrees S] as seen from the top.

    In this example, they used the cosine law with contained angle 115 degrees to find Fnet.

  • Physics -

    You can use the law of cosines here, but it is much more complicated. Two sides a,b are the vectors (head to tail), and c is the unknown side. The angle between the vectors is 180-60=120

    draw a triangle to prove that.

    c^2=a^2+b^2-2ab*cos120
    = 2*tension^2( 1-cos120)
    =2 tension^2 (1+cos60)
    = 2 tension^2(1+1/2)=3 tension^2
    c= tension*sqrt3 which is the same as above 2*tension*cos30.

    Now, with the law of cosines, you have then to use the law of sines to get the angle for the resultant.

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