Three spiders are resting on the vertices of a triangular web. The sides of the triangular web have a length of a = 0.85 m, as depicted in the figure. Two of the spiders (S1 and S3) have +7.7 µC charge, while the other (S2) has
−7.7 µC
charge.
a)What are the magnitude and direction of the net force on the third spider
(S3)
c)What are the magnitude and direction of the net force on the third spider (S3) when it is resting at the origin?
cant seem to get this one, what are the steps to do this?
F = k Q1 Q2/r^2
If all three sides are the same length a, we have an equilateral triangle with 60 degree interior angles
Q1=Q3 = Q
Q2 = -Q
find force on Q3
F due to Q1 in direction from Q1 through Q2 (repulsive)
F= k Q^2/a^2
F from Q2 on Q3 (attractive)
F= k Q^2 /a^2 same magnitude
so if x axis is along direction from Q1 to Q3 then
Fx = F - F cos 60
and if y is perpendiculat to x
Fy = F sin 60
Now move Q3 to the origin and do part c
so magnitude is (8.99E9*-7.7)/(.85)^2 ?
im still confused on distance?
To calculate the net force on the third spider (S3), we need to find the magnitudes of the forces between each pair of spiders and their respective directions.
First, let's find the force between S1 and S3, both charged with +7.7 µC:
F1 = k * Q1 * Q3 / r^2
F1 = (8.99E9) * (7.7E-6) * (7.7E-6) / (0.85)^2
F1 = 46.20 N
Next, let's find the force between S2 and S3, charged with -7.7 µC and +7.7 µC respectively:
F2 = k * Q2 * Q3 / r^2
F2 = (8.99E9) * (-7.7E-6) * (7.7E-6) / (0.85)^2
F2 = -46.20 N
Now, we need to find the directions of these forces. In this equilateral triangle, we can use 60° angles between the sides.
For F1 (between S1 and S3), the force is repulsive and directed radially outwards along the line between S1 and S3.
For F2 (between S2 and S3), the force is attractive and directed radially inwards along the line between S2 and S3.
To find the net force on S3, we can find the x and y components of the forces.
F1x = 0 (as it's along the line between S1 and S3)
F1y = F1 (as it's perpendicular to x-axis)
F2x = F2 * cos(60°) = -46.20 * 0.5 = -23.10 N
F2y = F2 * sin(60°) = -46.20 * 0.87 = -40.18 N
Now, we find the net force components:
Fx = F1x + F2x = 0 + (-23.10) = -23.10 N
Fy = F1y + F2y = 46.20 + (-40.18) = 6.02 N
Now, to find the magnitude and direction of the net force on S3:
F_net = sqrt(Fx^2 + Fy^2) = sqrt((-23.10)^2 + (6.02)^2) = 23.98 N
To find the direction, we can use the arctangent function:
θ = arctan(Fy / Fx) = arctan(6.02 / -23.10) = -14.42° (measured counterclockwise from the line connecting S1 and S3)
Now, when S3 is resting at the origin, the net force on S3 will have the same magnitude as the force between S1 and S2:
F_net (origin) = 46.20 N
The direction of the net force when S3 is at the origin will be directly towards the midpoint of the line connecting S1 and S2 (as both forces are attractive):
θ' = 60° (measured counterclockwise from the line connecting S1 and S3)
So for part (a):
Magnitude: 23.98 N
Direction: -14.42°
For part (c):
Magnitude: 46.20 N
Direction: 60°
To find the magnitude and direction of the net force on the third spider (S3), you can follow these steps:
1. Determine the distance between the spiders. Since all three sides of the triangle have a length of 0.85 m, the distance between each pair of spiders is also 0.85 m.
2. Use the formula for the force between charged particles: F = k * (Q1 * Q2) / r^2.
- Q1 represents the charge of the first spider (7.7 µC).
- Q2 represents the charge of the second spider (-7.7 µC).
- r represents the distance between the spiders (0.85 m).
- k is the electrostatic constant, approximately equal to 8.99 × 10^9 N·m^2/C^2.
3. Calculate the forces exerted on S3 by S1 and S2:
- F1 = k * (Q1 * Q3) / r^2, where Q3 represents the charge of S3 (7.7 µC).
- F2 = k * (Q2 * Q3) / r^2.
4. Determine the direction of each force:
- F1 is repulsive because both S1 and S3 have the same positive charge.
- F2 is attractive because S2 has a negative charge, attracting S3.
5. Calculate the net force on S3 by subtracting F2 from F1: F_net = F1 - F2.
Note: Magnitude and direction of net force can be determined separately.
For part c, where S3 is at the origin, you can follow these additional steps:
6. Determine the position of S3 at the origin. In this case, the distance between S3 and S1 or S2 is 0.85 m.
7. Repeat steps 3 and 4 to calculate the forces on S3 by S1 and S2, using the new distance (0.85 m).
8. Calculate the net force on S3 at the origin by subtracting the force exerted by S2 from the force exerted by S1: F_net_origin = F1 - F2.
By following these steps, you should be able to calculate the magnitude and direction of the net force on the third spider (S3) in both scenarios.
To find the magnitude and direction of the net force on the third spider (S3), we need to calculate the forces exerted by the other two spiders (S1 and S2) on S3, and then find the resultant force.
Step 1: Calculate the force exerted by S1 on S3 (F1).
The equation for the electrostatic force between two charges is given by F = k * (Q1 * Q2) / r^2, where F is the force, k is the Coulomb's constant (8.99 x 10^9 N*m^2/C^2), Q1 and Q2 are the charges, and r is the distance between the charges.
For S1 and S3, both have a charge of +7.7 µC.
Using the distance between them, which is the length of one side of the equilateral triangle (0.85 m), we can calculate F1:
F1 = (8.99 x 10^9 N*m^2/C^2) * (7.7 x 10^-6 C)^2 / (0.85 m)^2
Step 2: Calculate the force exerted by S2 on S3 (F2).
S2 has a charge of -7.7 µC.
Using the same distance as before, we can calculate F2:
F2 = (8.99 x 10^9 N*m^2/C^2) * (7.7 x 10^-6 C)^2 / (0.85 m)^2
Step 3: Find the components of forces F1 and F2.
Since the triangle is equilateral, the angle between the forces F1 and F2, relative to the x-axis, is 60 degrees.
The x-component of F1 is given by F1x = F1 - F1 * cos(60 degrees).
The y-component of F1 is given by F1y = F1 * sin(60 degrees).
Similarly, F2x = F2 * cos(60 degrees) and F2y = -F2 * sin(60 degrees).
Step 4: Calculate the net force on S3.
To find the net force, add the x-components and y-components of F1 and F2 separately.
Net force in the x-direction (Fx) = F1x + F2x.
Net force in the y-direction (Fy) = F1y + F2y.
The magnitude of the net force is given by the square root of the sum of the squares of the x-component and y-component:
Magnitude of net force = sqrt(Fx^2 + Fy^2).
To find the direction of the net force, you can use trigonometry to calculate the angle it makes with the x-axis:
Angle = atan(Fy / Fx).
To calculate the force on S3 when it is resting at the origin (part c), follow the same steps but use the new distances from the origin to each spider as the "r" value in the force equation.
I hope this explanation helps you understand the steps to solve the problem.