Friction
In the above figure it is known that when the blocks are released, they slide. Find the time (in seconds) when the small block will fall off the larger block.
Details and Assumptions
Assume the size of block of mass to be very small.
, , , ,
The answer is 2.
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1 solution
We know that: M=4 Kgr, m=1 Kgr, theta = tan(3/4)= 0.9316, mi=0.4 and l=4 m. Let g=10 m/s^2 , the accelaration a 1 of the larger block is : (M*g*sin(theta) -(M +m)*g*cos(theta)*mi)/M while the accelaration a 2 of the small block is: g sin(theta)-g cos(theta)* 0.5 mi. The small block has to cover a distance l + 0.5 * a_1 *t^2 in t seconds, so we get: l + 0.5 * a_1 *t^2 = 0.5 a 2 *t^2 or t = sqrt(2*l/(a 2 - a_1). We find that t is 2.1143 seconds or about 2 seconds.