CAN SOMEONE PLEASE CHECK THE LAST PART OF THIS QUESTION?
pete and john play a game of tug war on a frictionless icy curface. pete weighs 539 N and john weights 392 N. during the course of the game, john accelerates toward pete at a rate of 3 m/s^2. Sarah decideds to join the game as well. Now pete pulls on Sarah with a force of 45.0N [E] and john pulls on her with a force of 25N [N]. What's Sarah's resultant acceleration, if she weighs 294 N?
m=294N (this equals 30kg)
Fjohn=25 N [N]
Fpete=45 N [E]
Fnet=Fjohn+Fpete
=25^2 + 45^2
=51N
a=F/m
=51/30kg
=1.7m/s^2
Now, Im just confused about what angle to solve for. My diagram looks like this: have an arrowed line going west, to represent Pete and then from that arrow's head, I have an arrow going North to represent John. If this is correct, then I would solve for Tan, with the opposide side=25N and the the adjacent side equaling 45N, to give an angle of 29degrees. Is this correcct?
25 N and 45 E
tangent of angle above east axis = 25/45 = .556
Angle north of east = 29 deg
angle east of north = 90-29 = 61 deg
By the way I am sure you intended to say:
= sqrt (25^2 + 45^2)
To determine the resultant acceleration of Sarah, we need to consider the forces applied by Pete and John.
From the question, we know that Pete pulls on Sarah with a force of 45.0N to the east (positive x-direction), and John pulls on her with a force of 25N to the north (positive y-direction).
To find the resultant acceleration, we can use Newton's second law: Fnet = ma, where Fnet is the net force acting on Sarah, m is her mass, and a is the acceleration.
First, let's find the horizontal component (Fx) of the net force:
Fx = Fpete = 45N [E]
Next, let's find the vertical component (Fy) of the net force:
Fy = Fjohn = 25N [N]
Since the forces are at right angles to each other, we can treat Fx and Fy as the legs of a right triangle:
|
|
Fnet | 25N
|
------------------- ------------------------------
Fpete 45N
To find the net force (Fnet), we can use the Pythagorean theorem:
Fnet = sqrt(Fx^2 + Fy^2)
= sqrt((45N)^2 + (25N)^2)
= sqrt(2025N^2 + 625N^2)
= sqrt(2650N^2)
= sqrt(2650) N
≈ 51.48 N
Now we can calculate the acceleration (a):
a = Fnet / m
= 51.48 N / 30 kg
≈ 1.716 m/s^2
Therefore, Sarah's resultant acceleration is approximately 1.716 m/s^2.
Regarding the angle you mentioned, it seems like you were trying to find the angle between Fpete and Fjohn. However, in this question, we don't need to find that angle to determine Sarah's resultant acceleration.