Forces of 11.5 N north, 20.4 N east, and 15.7 N south are simultaneously applied to a 3.74 kg mass as it rests on an air table. What is the direction of the acceleration in degrees? (Take east to be 0 degrees and counterclockwise to be positive.)

Question

Forces of 11.5 N north, 20.4 N east, and 15.7 N south are simultaneously applied to a 3.74 kg mass as it rests on an air table. What is the direction of the acceleration in degrees? (Take east to be 0 degrees and counterclockwise to be positive.)

Answer 1

add the forces vectors.

a= F/m

combine the N-S forces, leaving you a S force, add that to E.

Answer 2

It's asking for the direction of acceleration. I calculated the acceleration already.

Answer 3

Use the two force vectors to get it.

YOu have a net S force, and a net E force.

tHe angle S of east is given by arctan S/E where S and E are the resultant vectors in those directions.

Answer 4

To find the direction of acceleration, we need to first find the net force acting on the mass. The net force is the vector sum of all the individual forces.

Given:
Force north (F_north) = 11.5 N, pointing in the y-direction (positive y-axis).
Force east (F_east) = 20.4 N, pointing in the x-direction (positive x-axis).
Force south (F_south) = 15.7 N, pointing in the negative y-direction.

Now, let's break down the individual forces into their x and y components:

F_north has no x-component and a positive y-component of 11.5 N.
F_east has a positive x-component of 20.4 N and no y-component.
F_south has no x-component and a negative y-component of 15.7 N.

Next, we can add up the x and y components of the forces separately to find the net force:

Net force in the x-direction (F_net_x) = F_east = 20.4 N (since it is the only force in the x-direction).

Net force in the y-direction (F_net_y) = F_north - F_south = 11.5 N - 15.7 N = -4.2 N.

So, the net force acting on the mass is F_net = F_net_x î + F_net_y ĵ, with F_net_x = 20.4 N and F_net_y = -4.2 N.

Now, let's find the magnitude and direction of the net force:

Magnitude of the net force (F_net) = √(F_net_x^2 + F_net_y^2)
= √((20.4 N)^2 + (-4.2 N)^2)
= √(416.16 N^2 + 17.64 N^2)
= √(433.80 N^2)
= 20.84 N (approximately)

Direction of the net force (θ_net) can be found using the inverse tangent function (tan^(-1)):

θ_net = tan^(-1)(F_net_y / F_net_x)
= tan^(-1)(-4.2 N / 20.4 N)
= tan^(-1)(-0.205)

Using a scientific calculator or trigonometric table, we find that tan^(-1)(-0.205) is approximately -11.71 degrees.

Since the question asks for the direction of acceleration, and acceleration is in the direction of the net force, the direction of acceleration in degrees is -11.71 degrees.