# The coefficient of static friction between the m = 3.50 kg crate and the 35.0° incline of the figure below is 0.270... ThanksWhat minimum force must be applied to the crate perpendicular to...

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

Thanks

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