00:01
Hello students, we are given a question that there is a diagram and in which there is a row which pulls both the crates in the right direction, okay, with a constant acceleration.
00:10
Now we are asked some questions or parts of the particular diagram and this activity.
00:15
First of all, we are asked that the type of force which acts on the 2kg crate, okay? see, this is 2kg crate, 2kg grade and the upper upper one is 1 .5 kggy crate and this one is the upper row.
00:30
This is in the black part which is totally connected with the 1 .5 kg grade now we are asked for the force which is which is a interaction pair to the forward force okay forward force is in the right direction and we are asked for the fbd fbd means is nothing but free body diagram and magnitude of the normal force which is acting on the 2kg grade and in e part we are given some friction between some slides okay and we are asked for the black backward force which acts on the 2kg crate and another part is asked that in ask is asked that there is a 1 .5 kg crate slips become because of the higher force suppose there is a if we pull the rope in a high pressure or can we said a high force obviously 1 .5 kg crate will go down and it will fall so basically what we are asked that what we are asked that what we are which forward force on the 2 kg crate and we are asked for the acceleration and the maximum tension on the proof now we will go with the first part okay what we are asked the type of force act on the 2kg crate we are supposed to know that the forward force which acts on the 2kg crate is it's a pure stetical frictional force okay because we know that the both the crates moves together okay in a static in a right direction with the static friction okay and that helps both crates to get moved so basically we can directly write as static frictional force okay and we have given the reason as well because we are supposed to know that both crates moves together in the right direction because there exists a static friction between them and it helps both the crates to get moved now we will come to the second part we are asked about which force which forces is in the interaction pair to the forward force okay we are now we are supposed to know that the interaction force is in the right direction and it will be in the interaction pair with the is the kinetic friction force okay it it will be in the two kg grade and the floor so we can directly write as kinetic frictional force okay kinetic frictional force now we will come our next part which is see in see what we are asked we are asked about the free body diagram okay so first of all we will make the free body diagram see how we will make it we can suppose that though we can directly say that the m2 is equal to 1 .5 kg okay and m1 is equal to 2 kg crates we are talking about crates okay and let's say n1 and n2 be the first at the second grade normal force okay first and second grade normal forces normal forces and one acts on the two kg grade and the n two credits on the 1 .5g grade okay now what we can do here is we can just make the diagram like that see how we will make the diagram we can make a 2 kg block okay it is nothing but m1 and we can make another block which can be considered as m2 it is 1 1 .5 kg grade we can say that the force will be acted by the row okay it is f and let's see we can say that there will be normal force it is and n2 and it is it will be m2g because of the gravity okay and we can say one more thing as well that it will be there will be as some kind of force which will be f limb okay we can say is say it as it as small f and where we can write it as the f limb or can be said of small f which can be considered as static friction force okay static friction force and we can say one more thing from here like n2 will be equal to m2g okay now what we can say from here that another part is we can draw one more diagram as well okay students and there will be let's say this is our block m1 and this is our another great block which is m2 okay and the same force will be act on consider as f limb we can write it as fs and there will be normal force n1 in the third and there will be the f limb f dash limb we can write it as f dash limb which can be considered as fk as well and fk is nothing but the kinetic friction force between the m1 and the floor okay and f s is nothing but the same static friction or force okay so we can write fs is the same okay and f k or can we said as f limp can be said as kinetic friction force okay students and we can say one more thing as well here n1 will be equal to see there are two blocks so we can say m1 plus m2 times of g it will be n1 it will be equal to m1 plus m to times of g it is our basic freebriety diagram now we will go to the fourth part what we are asked in the d part c we are given that the we are asked for the magnitude of the normal force which is acting on the two kg grade okay this is two kg grid we are asked for the normal force now what we can do here is we have observed in from the furibati diagram that n1 can be equal to m1 plus m2 times of g okay so what we can do from here we can do from here here that see n1 is nothing but m1 is nothing but two plus m2 is nothing but 1 .5 okay times we can consider g as direct 10 so it will be 3 .5 times 10 which comes out as 35 okay 35 newton now come to the e part we are given that see what we are given the data uk and us the value of uk and us it is the friction coefficient between the two crates okay and we are given the another friction coefficient which which is which acts on the 2 k grade and the floor we are asked for the backward force which acts on the 2kg crate okay now what we can do is here we can write as f of k is equal to u k times m1 plus m2 times of g okay so basically we can write it as fk is equals to see what is u k here f k comes out as kinetic friction between the m1 block 2 kg block and 2 kg crate and the and the floor okay so so what will with the uk? we can directly see that the 2kg crate and the floor.
07:43
So uk will be 0 .15.
07:46
Okay, we can write it as 0 .15 times of.
07:48
See, m1 plus m2 times of g is nothing but 35.
07:52
Okay, so we can write it as fk will be.
07:55
See, what will be the fk? it will be 5 .25...