00:01
There are two blocks we are given here so there are two blocks that is block a and block b.
00:09
These two things we are given here and the weight of the block a which is equal to we can write so if we talk about weight this is 1 kilo newton and the weight of the block b which is equal to we can write 2 kilo newton.
00:22
Now respectively they are in the equilibrium so these two blocks here both are here is in the equilibrium.
00:30
Now we talk about coefficient of the friction of the block so coefficient of friction of the two block as well as the block b in the floor which is equal to we can write 0 .13.
00:47
Find the minimum force so in this question we need to find minimum force p that required to move the block required to move the block b.
01:01
So let's solve this question and find the answer.
01:03
So here we can write the mass ma we can write 1 kilo newton here we can write mb we can write 2 kilo newton and that is mu a which is equal to mu b the coefficient of friction coefficient of friction which is equal to 0 .3.
01:23
These values we are given here.
01:24
Now we know that the component of the component here so component of the tension that is we can say t sin 30 degree to the t cos 30 degree.
01:39
So this we are given here this is a vertical component this is a horizontal component.
01:44
Now from this we can write free body diagram so we can here we can write summation of f y which is equal to 0.
01:50
So from the figure we can write n plus that is normal force acting on upward direction and plus we can write n.
01:57
We know that the mass that is capital mg on a downward side that is a t sin 30 degree which is equal to we can write and we know the capital m which is equal to ma here we can write ma plus we can write mb.
02:13
So here we can write ma plus mb.
02:16
Now again we just simplify and we can write n plus t whatever we can write sin 30 degree which is equal to ma plus mb which means both masses that is 1 kilo newton and 2 kilo newton which is equal to we can write 3 kilo newton per equation number 1.
02:31
Now summation of fx which is equal to 0 and here we can write the force which is equal to we can write minus t whatever we can write cos 30 degree plus we can write p and the friction that is a friction force.
02:48
So friction force which is equal to we can write f multiplied by mu that is substitute of value so mu multiplied by capital n which is equal to we can write minus t whatever we can write cos 30 degree and now we just simplify and we can write answer is capital n which is equal to we can write minus t whatever we can write cos 30 degree divided by we can write 0 .3 which is equal to we can write we just simplify and we can write answer is minus 2 .88675 multiplied by t plus we can write p.
03:17
This is our equation number 2.
03:19
So this is the free body diagram.
03:20
This will use our free body diagram on b here.
03:25
This is our using the free body diagram figure whole figure.
03:32
Now let's applying the free body diagram on a...