Revision Note 2: Static equilibrium

Conditions for static equilibrium:

1.       F_net = 0 (net force is zero)

2.       τ_net = 0 (net torque is zero)

For calculations:

·         Break F_net down into F_x, F_y (with proper signs) and then test for zeros along each axis.

·         For rotational moments, balance the clockwise and anti-clockwise moments.

Problem Solving Tips:

Tip 2.1:

In a force diagram, you will have to decide which direction is positive and which is negative. You can choose anything, but it helps to decide beforehand so that you do not waste time during the exam debating this.

Convention:

Vertically up is positive: E.g., g = -9.8 ms^-2.

Horizontally right is positive.

Along the Slope: x-axis, positive to the right.

Perpendicular to the slope: y-axis, positive upwards.

Tip 2.2:

Unknowns and signs: Suppose there is an unknown tension, T.

First, if you know its expected direction, draw it that way. Then use T or -T depending on the direction you chose. If you guessed the direction right, the numerical answer will be positive. If the numerical answer is negative, then the actual direction is in the reverse. It is like this: acceleration a = -5ms^-2 rightwards is actually +5ms^-2 leftwards.

Tip 2.3:

If θ is the angle of the inclined plane and a particle of mass m is on it, then the component of the weight along the x-axis and the y-axes are mg.sin θ and mg.cos θ respectively. Use the principle of extreme values to check this answer – when θ = 0 (plane is flat), then the component along the plane is 0 as it should be. This may seem counter to the “shadow = cos θ” tip, but is not since the θ is different. The correct angle (90 - θ) would match up with the shadow tip.

Tip 2.4: Up-down intuition:

Relating θ of inclined plane to a_x = component of g along the plane

θ ↑ a_x ↑

Intuition: When θ=90 then a_x = g, the highest possible value.