
Brian O. answered 01/18/16
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Unfortunately this part does not allow diagram or drawing.
Complete Solution can be provided on request.
Brian
Assuming the system is static i.e. no motion.
The system Free body diagram
The system is moving up or down depends on the strength of the two forces on a frictionless plane and they are M_1 g sin?θ and M_2 g
If the former is greater than the latter then object M1 is moving down the ramp and vice versa.
If frictional force is present, this force is always against the motion. For example, if the objects move down the ramp then the frictional force is up the ramp. So after solving (b) we know which direction are the objects moving we can then know the direction of the frictional force.
The two forces are: 7g sin 40 = 45 N and 60 N
M1 is moving up the ramp because M_1 g sin??θ<M_2 g?
Frictional force = M_1 gμ cos?θ = 7g*0.1 cos 40 = 5.36 N
The net force equation: M_2 g-M_1 g sin??θ-5.36=7a?
Where a is the acceleration
60 – 45 – 5.36 = 7a
a = 1.38 m/s2
Tension in the rope = M_2 g = 60 N
The system Free body diagram
The system is moving up or down depends on the strength of the two forces on a frictionless plane and they are M_1 g sin?θ and M_2 g
If the former is greater than the latter then object M1 is moving down the ramp and vice versa.
If frictional force is present, this force is always against the motion. For example, if the objects move down the ramp then the frictional force is up the ramp. So after solving (b) we know which direction are the objects moving we can then know the direction of the frictional force.
The two forces are: 7g sin 40 = 45 N and 60 N
M1 is moving up the ramp because M_1 g sin??θ<M_2 g?
Frictional force = M_1 gμ cos?θ = 7g*0.1 cos 40 = 5.36 N
The net force equation: M_2 g-M_1 g sin??θ-5.36=7a?
Where a is the acceleration
60 – 45 – 5.36 = 7a
a = 1.38 m/s2
Tension in the rope = M_2 g = 60 N