The coefficient of static friction between the m = 3.50 kg crate and the 35.0° incline of the figure below is 0.270...
What minimum force must be applied to the crate perpendicular to the incline to prevent the crate from sliding down the incline?
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The forces acting on the object of mas m on an inclined plane are :
mgcos (90-35) or mgsin35 along the surface down..........(1)
Normal force = mg*sin(90-35) +F....................(2)
Frictional force = Normal force *k along the inclined surface but opposite to the force at (1) Or (mgsicos35+F)k, where k =coeffcient of friction and is given to be 0.270
Therefore, (mg*cos35 +F)(-0.27) + mgsin35 = 0 for the equilibrium position, when the object is just prevented from being slid along. This is due our aplication of the minimum force F. Now we find F by solving for F:
0.27F = -mgsin35-0.27mgcos35
F = mg(sin35-0.27cos35)/(0.27)
=44.81422 Newton is the minimum required force to prvent the object from sliding down on the plane.
mass = m = 3.5 kg
Coefficient of friction = F = 0.270
Angle of incline = A = 35.0 degrees.
Acceleration due to gravity = g = 9.81 m/s^2
The force perpendicular to the incline due to the component of weight of crate =
f1 = m*g*Cos A = 3.5*9.81*Cos 35 = 3.5*9.81*0.8192 = 28.1272 N
The force parallel to the incline due to the component of weight of crate =
f2 = m*g*Sin A = 3.5*9.81*Sign 35 = 3.5*9.81*0.5736 = 19.694556 N
This force (f2) is pulling the crate down along the incline.
The frictional force preventing this movement =
f3 = f1*F = 28.1272*0.27 = 7.59435264 N
Additional frictional force required prevent the movement = f4 = f2 - f3 = 19.694556 - 7.59435264 = 12.10020336 N
Additional force vertical to incline required to provide this additional frictional force f4 = f5 = f4/0.270 = 44.815568 N
Minimum force required = 44.815568 N
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