Given: The friction between the block with mass 12 kg and the wedge with mass 16

Question

Given: The friction between the block with
mass 12 kg and the wedge with mass 16 kg
is 0.32 . The surface between the wedge with
mass 16 kg and the horizontal plane is smooth
(without friction).
The acceleration of gravity is 9.8 m/s2.
A block is released on the inclined plane
(top side of the wedge). The inclined plane is at an angle of 26 degrees with the horizontal.

What is the minimum force F which must
be exerted on the 16 kg block in order that the
12 kg block does not move down the plane?
Answer in units of N.

Explanation / Answer

Solution: mass of the block = 12 kg mass of the wdge = 16 kg friction b/w block and wedge , = 0.32 inclined plane is at an angle,= 26 degrees with the horizontal When the 12kg block is about to slip, the limiting frictional force on it as Fl= mgcos    = 0.32 x m x 9.8 x cos26o    = 2.818 m   (parllel to the slope)
component of this in the horizontal direction is         = (2.818 m)cos(26) = 2.532 m
horizontal component of friction, the 12kg block (m) must be accelerated horizontally at such a rate that the limiting frictional force is 'cancelled out'' by acceleration.
From newton's second law , F=ma, 2.532 m = m a acceleration of the block , a = 2.532 /s2
thus, minimum force F which must
be exerted on the 16 kg block in order that the
12 kg block does not move down the plane is Fmin = (12+ 16)( 2.532 ) =70.896 N

mass of the block = 12 kg mass of the wdge = 16 kg friction b/w block and wedge , = 0.32 inclined plane is at an angle,= 26 degrees with the horizontal When the 12kg block is about to slip, the limiting frictional force on it as Fl= mgcos    = 0.32 x m x 9.8 x cos26o    = 2.818 m   (parllel to the slope)
component of this in the horizontal direction is         = (2.818 m)cos(26) = 2.532 m
horizontal component of friction, the 12kg block (m) must be accelerated horizontally at such a rate that the limiting frictional force is 'cancelled out'' by acceleration.
From newton's second law , F=ma, 2.532 m = m a acceleration of the block , a = 2.532 /s2
thus, minimum force F which must
be exerted on the 16 kg block in order that the
12 kg block does not move down the plane is Fmin = (12+ 16)( 2.532 ) =70.896 N

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