Block a, with a mass of 10 kg, rests on a 30° incline. The coefficient of kinetic friction is 0. 20. The attached string is para
2 answers:
Consult the attached free body diagram.
Use Newton's second law to set up several equations:
• net force on block B
∑ F = F[tension] - (3.0 kg) g = (3.0 kg) a
Note that we take the positive direction to be "upward".
Equivalently,
F[tension] = (3.0 kg) a + 29.4 N
• net force on block A parallel to the incline
∑ F = (10 kg) g sin(30°) - F[tension] ± F[friction] = (10 kg) a
Note that we take "down the incline" to be the positive direction.
The sign of the frictional force depends on which direction block A moves along the incline. To determine which way the block will slide, suppose for a moment that there is no friction so that the equation reduces to
49 N - F[tension] = (10 kg) a
If we combine this equation with the one we got from block B, we eliminate F[tension] and find
(F[tension] - 29.4 N) + (49 N - F[tension]) = (3.0 kg) a + (10 kg) a
⇒ (13.0 kg) a = 19.6 N
⇒ a ≈ 1.5 m/s²
Since this acceleration is positive, we know block A wants to slide down the incline. Then F[friction] points in the negative direction (up the plane), so that
∑ F = (10 kg) g sin(30°) - F[tension] - F[friction] = (10 kg) a
or more simply,
49 N - F[tension] - F[friction] = (10 kg) a
and upon substituting F[tension] from above,
49 N - ((3.0 kg) a + 29.4 N) - F[friction] = (10 kg) a
49 N - (3.0 kg) a - 29.4 N - F[friction] = (10 kg) a
F[friction] = 19.6 N - (13 kg) a
• net force on A perpendicular to the incline
∑ F = F[normal] - (10 kg) g cos(30°) = 0
It follows that
F[normal] = (10 kg) g cos(30°) ≈ 85 N
so that
F[friction] = µ F[normal] ≈ 0.20 (85 N) = 17 N
Substitute F[friction] and solve for the acceleration a :
17 N ≈ 19.6 N - (13 kg) a
(13 kg) a ≈ 2.6 N
a ≈ 0.20 m/s²
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