__Tipping Force Calculator:__

__Tipping Force Calculator:__

Enter the values of mass of the magnet, m_{(kg)}, acceleration due to gravity, g(m/s^{2}), angle of the magnet, x and distance, a and b to determine the value of tipping force, TF_{(N)}.

__Tipping Force Formula:__

__Tipping Force Formula:__

Magnet tipping force is a measure of the force required to tip a magnet over an edge or pivot point, considering the gravitational pull on the magnet and the distribution of its mass relative to the pivot.

Tipping force, TF_{(N)} in Newton is calculated by dividing the product of mass of the magnet, m_{(kg)} in kilograms, acceleration due to gravity, g(m/s^{2}) in metres per second squared, angle of the magnet, x with the horizontal in degrees by the sum of distance, a and b in metres.

Tipping force, TF_{(N)} = m_{(kg)} * g_{(m/s2)} * cos(x) * b / (a+b)

TF_{(N)} = tipping force in Newton, N.

m_{(kg)} = mass of the magnet in kilogram, kg.

g_{(m/s2)} = acceleration due to gravity (approximately 9.81m/s^{2} on the surface of the Earth)

x = angle of the magnet with the horizontal in degrees.

a,b = distance in metres, m.

__Tipping Force Calculation:__

- Consider a rectangular magnet with a mass of 2 kg. The magnet is placed on a surface where one edge serves as a pivot point. The distance from the pivot to the center of mass is 0.1m, and the total length of the magnet is 0.2m, making b = 0.1m. If the magnet makes an angle of 30° with the horizontal, calculate the tipping force.

Given: m_{(kg)} = 2kg, g_{(m/s2)} = 9.81m/s^{2}, x = 30degree = pi/6, a = 0.1m, b = 0.1m.

Tipping force, TF_{(N)} = m_{(kg)} * g_{(m/s2)} * cos(x) * b / (a+b)

TF_{(N)} = 2 * 9.81 * cos(pi/6) * 0.1 / (0.1+0.1)

TF_{(N)} = 2 * 9.81 * ((3)^{1/2}/2) * 0.1 / 0.2

TF_{(N)} = 1.701 * (3)^{1/2} / 2

TF_{(N)} = 1.5N.

- The tipping force required to tip a magnet is 0.5 N, the distances from the pivot to the center of mass a is 0.1 metres, and from the center of mass to the edge opposite the pivot b is also 0.1 metres. The magnet is at an angle of 60degree with the horizontal. Calculate the mass of the magnet.

Given: TF_{(N)} = 0.5N, g_{(m/s2) }= 9.81m/s^{2}, x = 60degree = pi/3, a = 0.1m, b = 0.1m.

Tipping force, TF_{(N)} = m_{(kg)} * g_{(m/s2)} * cos(x) * b / (a+b)

m_{(kg)} = TF_{(N)} * (a+b) / g_{(m/s2)} * cos(x) * b

m_{(kg)} = 0.5 * (0.1+0.1) / 9.81 * cos(pi/3) * 0.1

m_{(kg)} = 0.5 * 0.2 / 9.81 * (1/2) * 0.1

m_{(kg)} =0.1 / 0.4905

m_{(kg)} = 0.2038kg.