Dynamics Problem

iamanoobatmath · #1 $Old$ March 3rd, 2009, 10:43 PM

Quote:

A mass of 25kg is being pulled to the right by an applied force of 85 N. A mass of 8 kg is located between the applied force and the 25 kg mass. The coefficient of friction is 0.2

Quote:

What is the friction force of each mass?

$(25*9.8)*0.2= 49 N$

$(8*9.8)*0.2=15.7 N$

So for the 25 kg mass the friction force is 49 N. The friction force for the 8 kg mass is 15.7 N

Quote:

What is the acceleration of each system?

$F=ma$

$Fa-Ff=ma$

$Fa=85$

$Ff=49+15.7=64.7$

$m=25+8=33$

$85-64.7=20.3$

$20.3=33a$

$\frac{20.3}{33}=a=0.615$

So the acceleration is 0.615 m/s^2

Quote:

What is the tension in the connecting rope?

The answer is supposed to be 64.4N however I do not know how to get this answer. Help!

mr fantastic · #2 $Old$ March 4th, 2009, 12:41 AM

Quote:

Originally Posted by iamanoobatmath $View Post$

$(25*9.8)*0.2= 49 N$

$(8*9.8)*0.2=15.7 N$

So for the 25 kg mass the friction force is 49 N. The friction force for the 8 kg mass is 15.7 N

$F=ma$

$Fa-Ff=ma$

$Fa=85$

$Ff=49+15.7=64.7$

$m=25+8=33$

$85-64.7=20.3$

$20.3=33a$

$\frac{20.3}{33}=a=0.615$

So the acceleration is 0.615 m/s^2

The answer is supposed to be 64.4N however I do not know how to get this answer. Help!

Well done showing your work (which I haven't actually checked but it gives the correct answer for tension - see below - so is probably OK).

Block B:

$F_{net} = T - \mu R = T - (0.20)(25g) = T - 49$ .... (1)

$F_{net} = ma = (25)(0.615) = 15.375$ .... (2)

Equate the above equations and easily solve for $T$ .