Question

. A block of mass M rests on a horizontal table.
Two weights with masses m1 = 50 Kg and m2 =
30 Kg, are attached to M as shown. The strings
are massless and the pulleys are frictionless. The
coecient of static friction between the block and
the table is mu s = 0.23. Find the minimum value
of M for which the arrangement is in a state of
static equilibrium.

Answer 1

draw a force diagram, work out components of each force in up and sideways directions.
im guessing the table is horizontal so you only need to look at the sideways forces. these will all add to give 0 when its in static equilibrium. Solve this for M

Answer 2

cant see the diagram so ive no idea how much of each force is in the horizontal direction

Answer 3

They gave a diagram. So do I use Fs, Fn, and Fg

Answer 4

i dont know what these forces relate to. Fs = force in the string?, Fn = force newton?

Answer 5

they just drew m1 on one side and m2 on another

Answer 6

here https://sakai.duke.edu/access/content/group/d873aea0-4598-4d59-9b46-2df4cbb23b39/Homework%20assignments/Homework2.pdf

Answer 7

okay so no angles
need your password to sign in, it should be fine without.

you want all the forces to balance

Answer 8

is it a table, the weights are attacked to M in the middle, and they hang of each side?

Answer 9

attached*

Answer 10

yes

Answer 11

okay so all the forces need to add to 0

F_m1 + F_m2 + F_M = 0

newtons law F = ma, F_m1 = 50g, F_m2 = -30g (im presuming its pointing right)
F_M = 0.23 x mg

g is gravitational constant = 9.81 m is the mass of block M

putting these into the first equation

50g -30g = mg

50 - 30 = m
m = 20

Answer 12

so that is how I can always find the minimum?

Answer 13

the minimum is when all the forces perfectly balance. they all add up to 0.

Answer 14

but then what about the coefficient?