ANOTHER QUICKQUESTION ABOUT THIS ONE, PLEASE HELP
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?
PHYSICS - Damon, Sunday, January 24, 2010 at 12:21pm
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
SO THE ANSWER IS 61 DEGREES? HAVE I GOT MY DIAGRAM SET UP WRONG THEN, BECAUSE I THOUGHT THAT I WAS SOLVING FOR THE TAN 25/45
Sarah needs to replace part of the metal railing on her deck and wants the new railing to match the existing railing. However, she does not know what metal the railing is made from. Which is a property that she can use to identify the metal?
To find Sarah's resultant acceleration, we need to consider the forces acting on her and then calculate the net force and divide it by her mass.
From the given information, John pulls Sarah with a force of 25 N [N], and Pete pulls her with a force of 45 N [E].
To determine the direction of the forces, we can consider the positive x-axis to be east, and the positive y-axis to be north.
The force from John is 25 N [N], which means it is acting in the positive y-direction (north).
The force from Pete is 45 N [E], which means it is acting in the positive x-direction (east).
Now, we can calculate the net force:
Fnet = Fjohn + Fpete
Fnet = 25 N [N] + 45 N [E]
To find the magnitude and direction of the net force, we can use the Pythagorean theorem and trigonometry. Let's calculate:
Magnitude: Fnet = sqrt((25 N)^2 + (45 N)^2)
Direction: The angle north of the east axis = arctan((25 N)/(45 N))
Angle east of north = 90 degrees - angle north of east
Using the given values, we get:
Magnitude: Fnet = sqrt((25)^2 + (45)^2) ≈ 51 N
Direction: Angle north of east = arctan(25/45) ≈ 29 degrees
Angle east of north = 90 - 29 degrees ≈ 61 degrees
Therefore, Sarah's resultant acceleration is approximately 1.7 m/s^2 in the direction 61 degrees east of north.