Adeli owner creates a lunch special display board by taking a uniform board which has a weight of 162 n, cutting it in half, hinging the halves together with a frictionless hinge, and setting it up as an inverted "v". determine the minimum coefficient of static friction needed between the board and the ground in order for her to set the display board up with an angle of 30° between the two sides.

The minimum coefficient of static friction is: 0.5

Explanation:

We need to represent the forces of the display board (see attached) as free body diagram. Then we need to remember that the display board is in translational and rotational equilibrium, so we can sum of forces acting to x axis and y axis getting from the sum of forces acting at x axis as:, we need to remember that friction force is: where u is the friction coefficient and N is the normal force at the surfaces in contact so replacing we get:, so this is the equation (1). Now we need to sum forces at y axis:, so . For doing the rotational equilibrium, we need to choose the ground as the point about which to take the torques: , this is for convenience because two forces passes through the point, as a result they don't have moment arm so we get:, so, this is the equation (2). Assuming that the board has a height as: L and remember that was cutting it in half, both sides have as longitude:L/2 and applying trigonometric identities, we can find as: and (see attached), and replacing this at equation (2), we get: and solving to Nboard we get. Now we can replace those results at equation (1), Now we can find the friction coefficient as:.

f=ma newton's second law 7x8=56n

electrical potential energy =0.072 j

explanation:

electrical potential energy is given by

u= q δv

q= charge=6 c

δv= potential difference and is given by

δv= e d

e= electric field

d= distance=5 mm=0.005 m

δv=2.4 (0.005)=0.012 v

now electric potential energy= u= q δv

u=6 (0.012)

u=0.072 j