I assume these are in a line.
friction= mu(M1+m2+m3)
netforce= totalmass*acceleration
104-mu(totalmass)=totalmass*acceleration
solve for totalmass, then solve for m2
if block2 is pushing block3,
forceonblock3-mu(m3)=m3*a
solve for forceonblock3
Three blocks rest on a frictionless, horizontal table, with m1 = 9 kg and m3 = 16 kg. A horizontal force F = 104 N is applied to block 1, and the acceleration of all three blocks is found to be 3.3 m/s2.
1) Find m2
2)What is the normal force between 2 and 3?
5 answers
I assume these are in a line.
friction= mu(M1+m2+m3)
netforce= totalmass*acceleration
104-mu(totalmass)=totalmass*acceleration
solve for totalmass, then solve for m2
if block2 is pushing block3,
forceonblock3-mu(m3)=m3*a
solve for forceonblock3
OOOPSsss. I forgot g as part of the weight. every where there is a mu, it should read mu*g*masses
friction= mu(M1+m2+m3)
netforce= totalmass*acceleration
104-mu(totalmass)=totalmass*acceleration
solve for totalmass, then solve for m2
if block2 is pushing block3,
forceonblock3-mu(m3)=m3*a
solve for forceonblock3
OOOPSsss. I forgot g as part of the weight. every where there is a mu, it should read mu*g*masses
what does u stand for in the above equations?
mu is the coefficent of friction...and it is given as zero, which is a very mythical situation. So you get to forget about friciton. Your teacher is too easy.
so what you're saying is:
forceonblock3-mu(m3)=m3*a
forceonblock3-0(m3)=...
forceonblock3=...
forceonblock3-mu(m3)=m3*a
forceonblock3-0(m3)=...
forceonblock3=...