Two blocks a and b have a weight of 10 lb and 6 lb, respectively. they are resting on the incline for which the coefficients of static friction are ma = 0.15 and mb = 0.25. determine the incline angle u for which both blocks begin to slide. also find the required stretch or compression in the connecting spring for this to occur. the spring has a stiffness of k = 2 lb> ft.

The angle to start the motion is 10.61°.

Compression in the spring=0.18 ft

Explanation:

Given that

coefficient of friction on block (μ)A=0.15

coefficient of friction on block(μ) B=0.25

mass of block A=10 lb

mass of block B=6 lb

Condition for begin the motion of blocks

If the component of gravity along incline plan will more than the friction force between surfaces than blocks will start to move.We know that friction will act opposite to the direction of motion so when blocks will move downward friction will act in opposite direction .

For block A

mg sinα =μmg cosα+F                   (F=2 x,x is the compression in the spring)

Where F is the spring force

Now by putting the values

[tex]10sinα=0.15(10)cosα+F

10 sinα=1.5 cosα+F--------1

For block B

mg sinα+F =μmg cosα

Now by putting the values

6 sinα+F=0.25 (6)cosα

6 sinα+F=1.5cosα------2

From equation 1 and 2

16 sinα=3 cosα

So α=10.61°

So the angle to start the motion is 10.61°.

Now putting the value of α any of equation to find spring compression

10 sinα=1.5 cosα+F

F=0.36 lb

0.36 lb=2 x

So compression in the spring=0.18 ft

ingenuity says that the answer is 2.7 m.