Find the minimum weight of block a in order for motion to be impending down the plane (to the left). assume the pulley to be frictionless. given: wb = 275 lbs, θ = 35 °, μ = 0.25 for all surfaces.

88.25lbs

Explanation:

Let N be the minimum weight to be found.

firstly we have to solve for the force body diagram for both blocks, for block B which is stable, we have its weight acting downwards,for it to move left affected by block A it would need a horizontal component which is the frictional force and we can calculate it as weight of B x coefficient of friction= 275lbs x 0.25 = 68.75lbs.

Calculating same for block A:(free body diagram)

the horizontal and vertical component of the forces on A would be:

Nsin35(horizontal comp) + Ncos35(0.25)(vertical comp)= 68.75lbs

=0.574N+0.205N=68.75lbs.

solving the above equation and making N subject of formula would result to 88.25lbs.

answer: 2×10^-2kg

explanation: density= mass per volume

vol=mass/density

converting to s.i units we have;

3÷1000kg = 0.003

1.5÷100m = 0.15

0.003/0.15 = 0.02kg