Question

The 50-lb block A and the 25-lb block B are supported by an incline that is held in the position.
Knowing that the coefficient of static friction is 0.15 between the two blocks and zero between block B and the incline, determine the value of θ for which motion is impending.

Answer 1

Explanation:

since motion is impends F= μN between A+B

Free body diagram of block A impending motion  

∑$F_{y} =0; N_{1} -50cos(theta)=0$

$N_{1} =50cos(theta)$

∑$F_{x}=0;T-50sin(theta) +$μ(1)$N_{1} =0$

$T=50sin(theta)$-μ(1)$(50)cos(theta)........(1)$

Free body diagram of block B impending motion

∑$F_{y} =0;N_{2}- N_{1} -25cos(theta)=0$

$N_{2}=75cos(theta)$

∑$F_{x}'\;T-$μ(1)$N_{1}-$μ(2)$N_{2} -25sin(theta)$

T=μ(1)(50)$cos(theta)+$μ(2)(75)$cos(theta)_25sin(theta)........(2)$

From Eq1 and Eq2

0=$25sin(theta)-$μ(1)$(100)cos(theta)-$μ(2)$(75)cos(theta)=0$

substituting in for μ(1)=0.15 and μ(2)=0 we have $15cos(theta)=25sin(theta);tan(theta)=15/25\\theta=31.0^o$