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!
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)>
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.
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.
To find the net force and direction of acceleration in this scenario, we need to consider the vector components of the forces exerted by both tugboats.
First, let's visualize the situation. Tugboat 1 is pulling in a direction given by [N 60 degrees E] and Tugboat 2 is pulling in a direction given by [S 60 degrees E].
To determine the net force, we need to break down the forces applied by each tugboat into their x and y-components.
For Tugboat 1:
- The force magnitude is 1.2 x 10^5 N, which remains the same for both tugboats.
- The angle given is 60 degrees. In this case, we consider the horizontal component (x-component) of the force, which would be adjacent to the angle.
- Using trigonometry, we can determine that the x-component of Tugboat 1's force is equal to (1.2 x 10^5 N) * cos(60 degrees).
For Tugboat 2:
- The force magnitude is also 1.2 x 10^5 N.
- The angle given is 60 degrees. In this case, we consider the vertical component (y-component) of the force, which would be opposite to the angle.
- Using trigonometry, we can determine that the y-component of Tugboat 2's force is equal to (1.2 x 10^5 N) * sin(60 degrees).
Now, we need to calculate the total net force. Since the two tugboats are pulling in different directions, we will add the x-components together and add the y-components together.
Fnet_x = (1.2 x 10^5 N) * cos(60 degrees) + (1.2 x 10^5 N) * cos(60 degrees)
Fnet_y = (1.2 x 10^5 N) * sin(60 degrees) + (1.2 x 10^5 N) * sin(60 degrees)
Now, let's simplify the expressions:
Fnet_x = (1.2 x 10^5 N) * cos(60 degrees) + (1.2 x 10^5 N) * cos(60 degrees)
= 2 * (1.2 x 10^5 N) * cos(60 degrees)
= 2.4 x 10^5 N * cos(60 degrees)
Fnet_y = (1.2 x 10^5 N) * sin(60 degrees) + (1.2 x 10^5 N) * sin(60 degrees)
= 2 * (1.2 x 10^5 N) * sin(60 degrees)
= 2.4 x 10^5 N * sin(60 degrees)
The net force can now be calculated using the Pythagorean theorem:
Fnet = sqrt(Fnet_x^2 + Fnet_y^2)
Using the given values:
Fnet = sqrt((2.4 x 10^5 N * cos(60 degrees))^2 + (2.4 x 10^5 N * sin(60 degrees))^2)
= sqrt((2.4 x 10^5 N)^2 * (cos^2(60 degrees) + sin^2(60 degrees)))
= sqrt((2.4 x 10^5 N)^2 * 1)
= 2.4 x 10^5 N
The direction of the net force can be found using trigonometry. The angle is given as 60 degrees in both cases, so we can say the net force is in the direction given by [E 60 degrees N].
Thus, the net force of the tugboats pulling the freighter is 2.4 x 10^5 N in the direction of [E 60 degrees N].