Question

4.64 relate force + motion => 2nd Law ?Fy2 = m2ay0 T - m2g = 0 T = (5.0kg)(9.8m/s²) T = 49N fr <= ?sN want equal for "breaking point" fr is up because acting against gravity ?Fx1 = m1ax0 T + ?sN - m1 g sin 35° = 0 49N + 0.25 (m1 g cos 35) - m1 g sin 35° = 0 m1 (9.8 m/s²) [sin 35° - 0.25 cos 35°] = 49N m1 = 49N / (3.6m/s²) = 14 kg ?Fy1 = m1ay0 N - m1 g cos 35 = 0 N = m1 g cos 35

          4.64 relate force + motion => 2nd Law
?Fy2 = m2ay0
T - m2g = 0
T = (5.0kg)(9.8m/s²)
T = 49N
fr <= ?sN want equal for "breaking point"
fr is up because acting against gravity
?Fx1 = m1ax0
T + ?sN - m1 g sin 35° = 0
49N + 0.25 (m1 g cos 35) - m1 g sin 35° = 0
m1 (9.8 m/s²) [sin 35° - 0.25 cos 35°] = 49N
m1 = 49N / (3.6m/s²) = 14 kg
?Fy1 = m1ay0
N - m1 g cos 35 = 0
N = m1 g cos 35

4.64 relate force + motion => 2nd Law
?Fy2 = m2ay0
T - m2g = 0
T = (5.0kg)(9.8m/s²)
T = 49N
fr <= ?sN want equal for "breaking point"
fr is up because acting against gravity
?Fx1 = m1ax0
T + ?sN - m1 g sin 35° = 0
49N + 0.25 (m1 g cos 35) - m1 g sin 35° = 0
m1 (9.8 m/s²) [sin 35° - 0.25 cos 35°] = 49N
m1 = 49N / (3.6m/s²) = 14 kg
?Fy1 = m1ay0
N - m1 g cos 35 = 0
N = m1 g cos 35

Added by Mary R.

University Physics with Modern Physics

Hugh D. Young 14th Edition

Instant Answer

Solved by Expert Sri K

01/19/2023

Step 1

To solve for x, use the quadratic equation: x2 + y2 = 9 Show more…

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relate force + motion => 2nd Law ΣFy2 = m2ay0 T - m2g = 0 T = (5.0kg)(9.8m/s²) T = 49N fr <= μsN want equal for "breaking point" fr is up because acting against gravity ΣFx1 = m1ax0 T + μsN - m1g sin 35° = 0 49N + 0.25 (m1 g cos 35) - m1 g sin 35° = 0 m1 (9.8 m/s²) [sin 35° - 0.25 cos 35°] = 49N m1 = 49N / (3.6m/s²) = 14 kg m1 g sin 35° T μsN m1 g cos 35° ΣFy1 = m1ay0 N - m1 g cos 35 = 0 N = m1 g cos 35 4.64