Considering a rectangular object of mass m 20 kg with width...

Considering a rectangular object of mass m = 20 kg, with width of 1.5m and height of 2m, on the ground. A horizontal force is applied on the object with a distance from the ground by h=0.9 m. Without considering tipping, determine the minimum magnitude of F that will cause the block to slide if $$ \mu_s = 0.25 $$: $$ F = $$ 1.5m F 2m h

Transcript

00:01 In this question we are given that we have a globe of mass 14 .2 kilogram that is resting on this horizontal layer.

00:13 A force f is acting at an angle of theta is given to us as 20 .8 degrees.

00:23 So we need to calculate the minimum value of f so that block starts moving.

00:28 Look, what is the frictional force acting on any object? fricksonal force is it is equal to destabilizing force.

00:47 Up to what point of time until unless it is destabilizing force is more than or equal to the mu static into n.

00:59 This is what f max, maximum frictional force available.

01:06 Once it is closed then what is the frictional force acting? it is mu kinetic into n.

01:13 Because before moving, the object will experiencing static friction.

01:19 So to move any object, we must apply a force at least equal to new static into n.

01:27 Over here, if you try to do the free -body diagram, the onward would be 142.

01:32 Upward would be this normal.

01:35 And if you break this f into two components, f -force 20 .8 and f -sign 20 .8.

01:44 This is the frictional force that is new static into n.

01:48 So in case of limiting conditions this normal would be equal to 1 142 minus f sign 20 .8 and new is what new static is 0 .55 0 .5 5 5 and 2 normal is 142 minus f sign 20 .8 that must be equal to f course 20 .8 right this comes from what sigma f vertical has to be 0.

02:26 This comes from sigma f horizontal has to be zero in the limiting condition.

02:32 All right.

02:34 This f is now required...

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