Two crewmen pull a raft through a lock, as shown in the figure . One crewman pulls with a force of 130 at an angle of 34° relative to the forward direction of the raft. The second crewman, on the opposite side of the lock, pulls at an angle of 45°. With what force should the second crewman pull so that the net force of the two crewmen is in the forward direction?
The answer must be in kN.
Below is how I worked out the problem, but it says my answer is wrong. Can someone please help?
130cos34° = 107.77
107.77/cos45° = 152.41
So, I thought .1524 kN was the answer. . .
Any suggestions? Thanks for any helpful replies!
Apparently you must use sine instead of cosine. Im still on a mission to find out why though.
To solve this problem, we can break down the forces into components. Let's assume the forward direction is the x-axis and the perpendicular direction is the y-axis.
First, let's find the x-component of the force applied by the first crewman:
F1x = 130 * cos(34°) = 107.396 kN (rounded to three decimal places)
Next, let's find the y-component of the force applied by the first crewman:
F1y = 130 * sin(34°) = 71.747 kN (rounded to three decimal places)
Now, let's find the x-component of the desired force applied by the second crewman so that the net force is in the forward direction. Since the two forces are in opposite directions along the x-axis, the x-components should add up to zero:
F2x + F1x = 0
F2x = -F1x
F2x = -107.396 kN (rounded to three decimal places)
Finally, let's find the y-component of the desired force applied by the second crewman so that the net force is in the forward direction. Since the two forces are in the same direction along the y-axis, the y-components should add up to obtain a net force of zero:
F2y + F1y = 0
F2y = -F1y
F2y = -71.747 kN (rounded to three decimal places)
Therefore, the magnitude of the force applied by the second crewman to achieve a net force in the forward direction is:
F2 = sqrt(F2x^2 + F2y^2)
= sqrt((-107.396)^2 + (-71.747)^2)
= sqrt(11526.733 + 5147.985)
= sqrt(16674.718)
≈ 129.146 kN (rounded to three decimal places)
So, the second crewman should pull with a force of approximately 129.146 kN.
To find the net force in the forward direction, you need to consider the horizontal components of the forces exerted by each crewman.
Let's calculate the horizontal component of the force exerted by the first crewman. The force of 130 N at an angle of 34° relative to the forward direction can be resolved into horizontal and vertical components using trigonometry. The horizontal component is given by:
Horizontal component = force * cos(angle)
= 130 N * cos(34°)
≈ 107.77 N
Now, let's calculate the horizontal component of the force exerted by the second crewman. The angle provided is 45°, but we must assume it is measured from the same reference direction as the angle for the first crewman (the forward direction). The horizontal component is given by:
Horizontal component = force * cos(angle)
= force * cos(45°)
= force * sqrt(2)/2
Since the net force should be in the forward direction, the horizontal components of both forces should add up to a positive value. Hence, the equation becomes:
107.77 N + force * sqrt(2)/2 > 0
To find the value of force that satisfies this inequality, isolate the variable:
force * sqrt(2)/2 > -107.77 N
force > -107.77 N * 2/sqrt(2)
force > -107.77 N * sqrt(2)
force > -152.41 N
Therefore, the second crewman should exert a force greater than 152.41 N (or approximately 0.15241 kN) in order for the net force to be in the forward direction.