An iron block of mass 10 kg is kept on a horizontal floor. The block is pulled by a rope at an angle `30^(@)` with the floor. What should be the minimum force necessary to set the block in motion. Given `mu` = 0.5

Explanation

Let the minimum force applied by the rope be T which makes an angle `theta` with the horizontal. Horizontal and vertical components of T are T cos `theta` and T sin `theta` respectively. T sin `theta ` acts in the direction of the normal force ( R) as shown in the diagram . In this problem, `theta = 30 ^(@)` .

As the block is about to move the frictional force becomes equal to the force of limiting friction. The resultant of forces acting on the body is zero as the block is still stationary.
According to the figure, net horizontal force = T cos `theta - mu` R
net vertical force = R + T sin `theta `-W
As the block is stationary,
`:." ""T cos" theta -mu R = 0 " ""or", mu R = T "cos" theta`
and R+Tsin `theta` -W = 0`" ""or",` R = W -T sin `theta`
Dividing (1) by (2) , we get , `mu = (Tcostheta)/(W-T sintheta)`
or, `" " T= (muW)/(musintheta+costheta)=(0.5xx10xx9.8)/(0.5xx(1)/(2)+(sqrt(3))/(2))`
`=(49)/(0.25+0.866)`=43.9N