The coefficient of static friction between the *m* = 3.50 kg crate and the 35.0° incline of the figure below is 0.270...

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What minimum force must be applied to the crate perpendicular to the incline to prevent the crate from sliding down the incline?

Figure:

http://www.webassign.net/sercp/p4-41alt.gif

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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

=mg(sin35-0.27cos35) Or

F = mg(sin35-0.27cos35)/(0.27)

= 3.5*9.81(sin35-0.27cos35)/(0.27)

=44.81422 Newton is the minimum required force to prvent the object from sliding down on the plane.

Given:

mass = m = 3.5 kg

Coefficient of friction = F = 0.270

Angle of incline = A = 35.0 degrees.

Assumed:

Acceleration due to gravity = g = 9.81 m/s^2

Solution:

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

Answer:

Minimum force required = 44.815568 N

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