Ross and Rachel are fighting if they were on a break. Ross is trying to get his stuff weighing 150. N placed on an inclined plane by pulling it upwards parallel to the inclined plane with 150. N force. However, Rachel is trying to stop Ross by exerting 50.0 N perpendicular to the inclined plane. The coefficients of friction between the box and the incline are = 0.450 and y=0.350. Which of the following shows the correct equilibrium equation in the direction parallel to the inclined surface? 50 N FRIENDS A. EF = 150- W sin(30) - f + 50sin (30) = 0 B. EF = 150-W cos(30) + N - f = 0 C. EF = 150-W cos(30) - f = 0 D. EF = 150-W sin(30) - f = 0 150 N

Structural Analysis
6th Edition
ISBN:9781337630931
Author:KASSIMALI, Aslam.
Publisher:KASSIMALI, Aslam.
Chapter2: Loads On Structures
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Ross and Rachel are fighting if they were on a break. Ross
is trying to get his stuff weighing 150. N placed on an
inclined plane by pulling it upwards parallel to the inclined
plane with 150. N force. However, Rachel is trying to stop
Ross by exerting 50.0 N perpendicular to the inclined
plane. The coefficients of friction between the box and the
incline are = 0.450 and Uk=0.350.
Which of the following shows the correct equilibrium
equation in the direction parallel to the inclined surface?
50 N
FRIENDS
A. EF = 150- W sin(30) - f + 50sin (30) = 0
B. EF = 150-W cos(30) + N - f = 0
C. EF =
150-W cos(30) - f = 0
D. EF = 150W sin(30) - f = 0
150 N
Transcribed Image Text:Ross and Rachel are fighting if they were on a break. Ross is trying to get his stuff weighing 150. N placed on an inclined plane by pulling it upwards parallel to the inclined plane with 150. N force. However, Rachel is trying to stop Ross by exerting 50.0 N perpendicular to the inclined plane. The coefficients of friction between the box and the incline are = 0.450 and Uk=0.350. Which of the following shows the correct equilibrium equation in the direction parallel to the inclined surface? 50 N FRIENDS A. EF = 150- W sin(30) - f + 50sin (30) = 0 B. EF = 150-W cos(30) + N - f = 0 C. EF = 150-W cos(30) - f = 0 D. EF = 150W sin(30) - f = 0 150 N
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