Chapter+5+-+STATICS

__**Chapter 5 - STATICS**__ //__**Prior Knowledge Required:**__// Resolving Forces:
 * When resolving forces, resolve both parallel and perpendicular to the plane
 * The reaction always acts perpendicular to the plane.....NOT vertically upwards
 * Don't forget to take the correct components of forces that are not acting in the same direction as the plane



Given that the angle to the horizontal is 30, find the coefficient of friction (u) given that it is held in limiting equilibrium Resolving parallel to the plane: F = 30gsin30 Resolving perpendicular to the plane: R=30gcos30 Since the particle is in limiting equilibrium, F=uR, 30gsin30=u30gcos30 Therefore, u = tan30 = 1/rt3 = 0.577 (3sf)

Moments: A moment is the ability of a force, to turn the object it is acting on. It is calculated by "force x perpendicular distance" In statics, as there is no movement, we say that the anticlockwise (ACW) moments must equal the clockwise (CW) moments. To determine whether a moment is CW or ACW, imagine attaching a piece of string to the point where the force acts, and 'pull' in the direction of the force. Then it is easier to tell whether the force causes the system to move CW or ACW about a given point.



Find an expression for the weight, 30g, in terms of R and S. The length of the diagonal is 3y, and the distance between T and B is 2y. It is generally best to take moments about the point at which most forces act. In this case, that would be at C where both a reaction T and a resistive force F act. About C, S and R are clockwise moments and the weight, 30g, is anticlockwise. CW = ACW S(3ysin60) + R(2y) = 30g(2ycos60) Note that I have bracketed the distance to the point C so that the method is clearer. (rt3)Sy/2 + 2Ry = 30gy (rt3)Sy/2y + 2Ry/y = 30g Therefore, 30g = (rt3)S/2 + 2R

Typical Style Exam Questions:

Resolving forces horizontally: F = S and vertically: R = 5W

Since there is limiting equilibrium, u takes its maximum value, such that F=uR Therefore, F = 5uW and hence, S = 5uW

Taking moments about A (since it eliminates more unknowns): CWM = ACWM 2Wacosθ + 4W(3acosθ) = S(4asinθ) Replacing S for 5uW and cancelling the W and a: 2cosθ + 12cosθ = 20usinθ Dividing by cosθ: 14 = 20utanθ Since tanθ = 2 14 = 40u and therefore, u = 0.35 as required