Help please cylinder, torque, rolling w/o slipping problem

[nchin]

A massless string is wrapped around a 10 kg cylinder. the picture will look like the letter b. you pull upward on the string with a force of P = 49N. The bottom of the cylinder s on horizontal table and will roll w/o slipping. using dynamics involving center of mass, find the magnitude and the direction of frictional force (f). (no slippage between string and cylinder)Solution:

Torque about cm: RP - Rf = Iα ---> R(P-f) = Iα

I = 1/2MR^(2)
α = a/R

R(P-f) = 1/2MR^(2) (a/R) ---> P-f = 1/2Ma

Forces Horz: f = ma

Why do we need to consider the forces horizontal?? Isn't it already included in the torque calculation?

P-ma = 1/2Ma
P = 3/2Ma
Ma = 2/3P
a = 2/3 (49N) / 10kg ---> 3.2667 N

My teacher got 32.667N? but how?

 

[SammyS]

A massless string is wrapped around a 10 kg cylinder. the picture will look like the letter b. you pull upward on the string with a force of P = 49N. The bottom of the cylinder s on horizontal table and will roll w/o slipping. using dynamics involving center of mass, find the magnitude and the direction of frictional force (f). (no slippage between string and cylinder)


Solution:

Torque about cm: RP - Rf = Iα ---> R(P-f) = Iα

I = 1/2MR^(2)
α = a/R

R(P-f) = 1/2MR^(2) (a/R) ---> P-f = 1/2Ma

Forces Horz: f = ma

Why do we need to consider the forces horizontal?? Isn't it already included in the torque calculation?

You have one equation with two unknowns. You need another equation in the same two unknowns.

f = Ma gives you that.

P-ma = 1/2Ma
P = 3/2Ma
Ma = 2/3P
a = 2/3 (49N) / 10kg ---> 3.2667 N

My teacher got 32.667N? but how?

You solved for a and got 3.2667 m/s² . (You had the wrong units.)

What does that give you for f ?

 

[nchin]

You have one equation with two unknowns. You need another equation in the same two unknowns.

f = Ma gives you that.


You solved for a and got 3.2667 m/s² . (You had the wrong units.)

What does that give you for f ?

ooh ok i got it. thanks!