Friction = uN
N=mg so
F=umg
F at t(0) = (.505)(4.17 kg)(9.8 m/s^2)
F at t>0 = (.255)(4.17 kg)(9.8 m/s^2)
A 4.170 kg block of wood rests on a steel desk. The coefficient of static friction between the block and the desk is μs = 0.505 and the coefficient of kinetic friction is μk = 0.255. At time t = 0, a force F = 12.7 N is applied horizontally to the block. State the force of friction applied to the block by the table at the following times:
t=0 and t>0
3 answers
In both cases, the force of friction is 12.7 N because the box won't move. It's just being pushed once and halts. Hope this helps:
𝑓≤𝜇s𝑁
This means that the force of static friction can take on any value from 0
to 𝜇𝑁. Normal force = mg
static force vector=[0,𝜇s𝑚𝑔]
= [0,0.505×(4.17 kg)×(9.81 m/s2)] = [0,20.7]
At time 𝑡=0 the applied force is 𝐹=12.7 N, well below the maximum static frictional force which is 20.7 N. Therefore, the frictional force can and will rise to match this external force. Since the object remains motionless, the frictional force can maintain this value indefinitely, that is, for 𝑡>0.
𝑓≤𝜇s𝑁
This means that the force of static friction can take on any value from 0
to 𝜇𝑁. Normal force = mg
static force vector=[0,𝜇s𝑚𝑔]
= [0,0.505×(4.17 kg)×(9.81 m/s2)] = [0,20.7]
At time 𝑡=0 the applied force is 𝐹=12.7 N, well below the maximum static frictional force which is 20.7 N. Therefore, the frictional force can and will rise to match this external force. Since the object remains motionless, the frictional force can maintain this value indefinitely, that is, for 𝑡>0.
Therefore, the force of friction applied by the table on the block is 12.7 N at both times t=0 and t>0.
0 answers