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Sagot :
The largest angle for which box B will not slip is 16.7⁰.
The given parameters:
- weight of box, W = 10 lb
- coefficient of static friction between box A and box B, [tex]\mu_s_1[/tex] = 0.24
- coefficient of static friction between box B and the inclined surface, [tex]\mu_s_2[/tex] = 0.3
The frictional force between box A and box B is calculated as;
[tex]F_f_1= \mu_s_1 (mg)[/tex]
The frictional force between box B and inclined surface is calculated as;
[tex]F_f = \mu_s_2 (2mg) cos(\theta)[/tex]
The net force on the boxes that results in no movement is calculated as;
[tex](2mg) sin(\theta) - F_f= 0\\\\(2mg) sin(\theta) = F_f\\\\(2mg) sin(\theta) = \ \mu_s_2 (2mg) cos(\theta)\\\\ sin(\theta) = \mu_s_2 cos(\theta)\\\\\mu_s_2 = \frac{sin(\theta)}{cos(\theta)} \\\\\mu_s_2 = tan(\theta)\\\\\theta = tan^{-1} (\mu_s_2)\\\\\theta = tan^{-1} (0.3)\\\\\theta = 16.7 \ ^ 0[/tex]
Thus, the largest angle for which box B will not slip is 16.7⁰.
Learn more about coefficient of static friction here: https://brainly.com/question/25050131
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