A flea pulls on a puck with force 7 microN at an angle of 30°. Another flea pulls on the puck with force 2 microN at an angle of -60°.
a) What is the magnitude of the net force on the puck? (I got 7.28 uN)
b)At what angle is the net force? ( I got about 14.07 deg)
c) A third flea decides to pull the puck so that it will stop accelerating. What is the magnitude of the force with which it must pull, and in what direction must the flea pull so that the puck stops accelerating?
This is the one I need help with. Is it similar to part a?
a) sqrt(2^2+7^2) = sqrt(53) = 7.28 check
b)tan^-1 (2/7) = 15.9 and 30-15.9 = 14.1 check
c) you need the third flea to cancel the RESULTANT of a and b
He must pull at 7.28 *10^-6 N opposite to part b which is 14.1 + 180 = 194 degree(or 14.1 - 180)
It depends on how the angle is measured. If the angle is measured from some arbritary axis, and the fleas are on the oppsite sides of this axis, then you have a ninety degree angle between the force vectors
You are correct on a, b. On C, the equilibrant force is equal and opposite to a.
I see it as 7.28uN at the angle opposite b.
Ah, good, I am taking a break :)
To solve part c, we need to determine the magnitude and direction of the force required to stop the puck from accelerating. This can be done using vector addition.
First, let's calculate the x and y components of the two forces from parts a and b:
Force 1:
Magnitude = 7 μN
Angle = 30°
Force 2:
Magnitude = 2 μN
Angle = -60°
To find the x and y components, we'll use trigonometric functions. The x-component of a force can be found using the formula: Force * cos(angle), and the y-component can be found using: Force * sin(angle).
For Force 1:
Fx1 = 7 μN * cos(30°) ≈ 6.06 μN
Fy1 = 7 μN * sin(30°) ≈ 3.50 μN
For Force 2:
Fx2 = 2 μN * cos(-60°) ≈ 1 μN
Fy2 = 2 μN * sin(-60°) ≈ -1.73 μN
Now, let's calculate the net force in the x and y directions:
Net Fx = Fx1 + Fx2 ≈ 6.06 μN + 1 μN ≈ 7.06 μN
Net Fy = Fy1 + Fy2 ≈ 3.50 μN + (-1.73 μN) ≈ 1.77 μN
To find the magnitude of the net force, we can use the Pythagorean theorem:
Net force magnitude = sqrt(Net Fx^2 + Net Fy^2) ≈ sqrt((7.06 μN)^2 + (1.77 μN)^2) ≈ 7.28 μN (same as in part a)
Now, let's find the direction of the net force. We can use the inverse tangent function (tan^(-1)):
Angle of the net force = tan^(-1)(Net Fy / Net Fx)
Angle of net force ≈ tan^(-1)(1.77 μN / 7.06 μN) ≈ 14.07° (same as in part b)
Now, to determine the force and direction required to stop the puck from accelerating, we need to find the force that will cancel out the net force in part c.
The magnitude of the force required to stop the puck's acceleration will be the same as the magnitude of the net force: 7.28 μN.
To determine the direction in which the third flea must pull, we need to find the angle opposite to the net force angle. Considering that the net force angle is 14.07°, the third flea should pull in the opposite direction, which is 180° out of phase. Therefore, the third flea should pull with a force of 7.28 μN at an angle of 180 + 14.07 = 194.07°.
So, the magnitude of the force required to stop the puck's acceleration is 7.28 μN, and the direction in which the flea must pull is 194.07°.