Two workers are trying to move a heavy crate. One pushes on the crate with a force A, which has a magnitude of 445 N and is directed due west. The other pushes with a force B, which has a magnitude 325 N and is directed due north. What is the direction that the workers push the crate with respect to North?
tanAr = Y/X = 325/-445 = -0.73034
Ar = -36o = Reference angle.
A = -36 + 180 = 144o CCW = 54o W. of N.
To determine the direction that the workers push the crate with respect to North, we can find the resultant force by adding the individual forces exerted by each worker.
First, let's represent the forces A and B as vectors:
A = -445 N (due west)
B = 325 N (due north)
Now, let's find the resultant force of A and B. To do this, we need to find the horizontal and vertical components of each force.
Horizontal component of A (Ax) = A * cos(90) = -445 * cos(90) = -445
Vertical component of B (By) = B * sin(0) = 325 * sin(0) = 0
Since the vertical component of B is zero, it has no effect on the crate's direction.
The resultant force (R) can be found by summing the horizontal and vertical components:
Rx = Ax = -445 N
Ry = By = 0
To find the magnitude and direction of R, we can use the Pythagorean theorem and trigonometry:
R = sqrt(Rx^2 + Ry^2) = sqrt((-445)^2 + 0^2) = 445 N
The direction of R with respect to North can be found using the inverse tangent function:
θ = tan^(-1)(Ry / Rx) = tan^(-1)(0 / -445)
θ = tan^(-1)(0) = 0°
Therefore, the direction that the workers push the crate with respect to North is 0°, or due north.
To find the direction in which the workers push the crate with respect to North, we need to determine the resultant force vector. This can be done by using vector addition.
To add the vectors A and B, we can treat them as horizontal and vertical components of a single vector. The horizontal component due to force A is -445 N (directed west) and the vertical component due to force B is +325 N (directed north). Since north is considered the positive y-axis direction, we can write the vector sum as:
Resultant force vector = -445 N î + 325 N ĵ
Now, we can find the angle that the resultant force vector makes with the positive y-axis (North) by using trigonometry. The angle θ can be found using the inverse tangent function:
θ = tan^(-1)(vertical component / horizontal component)
= tan^(-1)(325 N / -445 N)
Calculating this in degrees, we find:
θ ≈ tan^(-1)(-325/445) ≈ -36.8°
Since the angle is negative, it means the resultant force vector is directed in the fourth quadrant (southwest) with respect to North.
Therefore, the workers push the crate in a direction approximately 36.8° south of due west with respect to North.