Question

Four blocks are connected by a weightless cord. Block A hangs from a table on the left. Block B and C rest upon the table. Block D hangs from the table on the right. Block A and D are over pulleys. Block B has a weight of 1N, Block C=4N and Block D=7N.

a. Find the weight of Block A such that the accelartion of Block B is (g/7)

b. Find the coefficient of friction for the table, if the acceleration of block C is (g/14)

c. Find the amount of mass that must be added to B and C to have the acceleration of Block D to be (g/21) if the coeeficient of friction is its max in B.

Answer 1

part a:

let weight of block A is F N.

let tension in the cord connecting B and C be T1 N, tension in the cord connecting C and D be T2 N and tension in the cord connecting A and B be T N.

writing force equation for D:

weight of D-tension in the cord connecting between C and D=mass of D*acceleration of D

==>7-T2=(7/g)*(g/7)=1

==>T2=6 N

writing force equation for C:

T2-T1=(4/g)*(g/7)=4/7

==>T1=6-(4/7)=5.42857

writing force equation for B:

T1-T=(1/g)*(g/7)

==>T=5.2857 N

then writing force equation for A:

T-weight of A=mass of A*(g/7)=weight of A/7

==>weight of A=T/(1+(1/7))=4.625 N

part b:

let coefficient of friction be k.

equation for D will be:
7-T2=(7/g)*(g/14)
==>T2=6.5 N

equation for C:
T2-T1-friction force=(4/g)*(g/14)

==>6.5-T1-k*4=4/14

==>T1=6.21428-4*k....(1)

equation for B:

T1-T-friction force=(1/g)*(g/14)

==>6.21428-4*k-T-k*1=1/14

==>T=6.142857-5*k

writing equation for A:

T-4.625=(4.625/g)*(g/14)

==>6.142857-5*k-4.625=0.330357

==>k=0.2375

hence coefficient of friction is 0.2375 .

part c:

let new weight to be added is F N.

then weight of B=1+F N

weight of C=4+F N

then force equation for D:

7-T2=(7/g)*(g/21)

==>T2=6.67 N

force equation for A:

T-4.625=(4.625/g)*(g/21)

==>T=4.8452 N

force equation for B:

T1-T-friction force=mass*acceleration

==>T1-4.8452-(1+F)*0.2375=(1+F)/g)*(g/21)=(1+F)/21

==>T1=4.8452+0.28512*(1+F)=5.13032+0.28512*F...(2)

force equation for C:

T2-T1-friction force=mass *acceleration

==>6.67-(5.13032+0.28512*F)-0.2375*(4+F)=((4+F)/g)*(g/21)

==>6.67-5.13032-0.28512*F-0.95-0.2375*F=0.190476+0.04762*F

==>F=(6.67-5.13032-0.95-0.190476)/(0.28512+0.2375+0.04762)=0.7 N

mass to be added=F/g=0.071435 kg