Question

B C a_r = -0.5195 m/s² a_? = -0.3 m/s² r 600 mm ? $\sum$ F_r = ma_r A 600 mm ? W = 3(9.81) 30° r 30° N F N cos 30° - 3(9.81) sin 30° = 3(-0.5195) N = 15.19 N $\sum$ F_? = ma_? F + 15.19 sin 30° - 3(9.81) cos 30° = 3(-0.3)

          B
C
a_r = -0.5195 m/s²
a_? = -0.3 m/s²
r
600 mm
?
$\sum$ F_r = ma_r
A
600 mm
?
W = 3(9.81)
30°
r
30°
N
F
N cos 30° - 3(9.81) sin 30° = 3(-0.5195)
N = 15.19 N
$\sum$ F_? = ma_?
F + 15.19 sin 30° - 3(9.81) cos 30° = 3(-0.3)

B
C
ar = -0.5195 m/s²
a? = -0.3 m/s²
r
600 mm
?
∑ Fr = mar
A
600 mm
?
W = 3(9.81)
30°
r
30°
N
F
N cos 30° - 3(9.81) sin 30° = 3(-0.5195)
N = 15.19 N
∑ F? = ma?
F + 15.19 sin 30° - 3(9.81) cos 30° = 3(-0.3)

Added by Scott G.

University Physics with Modern Physics

Hugh D. Young 14th Edition

Instant Answer

Solved by Expert Lucas Finney

Step 1

First, we have the equation 600 mm = ma N cos30 - 39.81 sin30 = 3 - 0.5195. This equation represents the sum of forces in the x-direction. The left side of the equation represents the force due to acceleration (ma), and the right side represents the force due to Show more…

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Can you please explain how we determine whether it's sin or cos on each axis for each component? Bar = -0.5195 m/s^2 a = -0.3 m/s^2 600 mm = ma N cos30 - 39.81 sin30 = 3 - 0.5195 600 mm N = 15.19 N W = 39.81 * 30 = mag F + 15.19 sin30 - 39.81 cos30 = 3 - 0.3 30