- #1
ubiquinone
- 43
- 0
Hi, I 'm just reviewing some forces problem. I have this one question that I was wondering if anyone could please clarify for me. Thank you.
Question: Block [tex]B[/tex] is initially at rest on block [tex]A[/tex], which is at rest on a smooth frictionless floor. There is friction between block [tex]A[/tex] and block [tex]B[/tex]. The coefficient of static friction between the blocks is [tex]0.200[/tex] and the coefficient of kinetic friction between the blocks is [tex]0.150[/tex]. [tex]m_A=6.00kg[/tex] and [tex]m_B=4.00kg[/tex]. A horizontal force of [tex]F=23.0N[/tex] is applied to block [tex]B[/tex].
Calculate the magnitude of the frictional force on block [tex]B[/tex] showing wheteher [tex]B[/tex] slides on [tex]A[/tex] or not.
In order to find out whether block [tex]B[/tex] would slide on block [tex]A[/tex], I need to find out if [tex]F_{fmax}[/tex] is less than or greater than the horizontal force [tex]F[/tex] applied on block [tex]B[/tex]
However, how can I find [tex]F_{fmax}[/tex]? Do I just use the formula [tex]F_{fmax}=\mu_sm_Ag[/tex]
I think it should be block A because the normal force of mass A is pushing block B up.
Question: Block [tex]B[/tex] is initially at rest on block [tex]A[/tex], which is at rest on a smooth frictionless floor. There is friction between block [tex]A[/tex] and block [tex]B[/tex]. The coefficient of static friction between the blocks is [tex]0.200[/tex] and the coefficient of kinetic friction between the blocks is [tex]0.150[/tex]. [tex]m_A=6.00kg[/tex] and [tex]m_B=4.00kg[/tex]. A horizontal force of [tex]F=23.0N[/tex] is applied to block [tex]B[/tex].
Calculate the magnitude of the frictional force on block [tex]B[/tex] showing wheteher [tex]B[/tex] slides on [tex]A[/tex] or not.
In order to find out whether block [tex]B[/tex] would slide on block [tex]A[/tex], I need to find out if [tex]F_{fmax}[/tex] is less than or greater than the horizontal force [tex]F[/tex] applied on block [tex]B[/tex]
However, how can I find [tex]F_{fmax}[/tex]? Do I just use the formula [tex]F_{fmax}=\mu_sm_Ag[/tex]
I think it should be block A because the normal force of mass A is pushing block B up.