A firefighter of mass 101 kg slides down a vertical pole with an acceleration of 3.2 m/s2. What is the friction force that acts on the firefighter?
+33665 Here we have net force = mass*acceleration.
net force = force of gravity - frictional force (assuming any other forces are negligible).
Assuming the frictional force is constant throughout his slide, we have:
101*9.81 - fricforce = 101*3.2 (where gravitational acceleration = 9.81m/s2)
rearrange to get fricforce:
fricforce = 101*9.81 - 101*3.2 = 101*6.61 N
$${\mathtt{fricforce}} = {\mathtt{101}}{\mathtt{\,\times\,}}{\mathtt{6.61}} \Rightarrow {\mathtt{fricforce}} = {\mathtt{667.61}}$$ Newtons
+33665 Here we have net force = mass*acceleration.
net force = force of gravity - frictional force (assuming any other forces are negligible).
Assuming the frictional force is constant throughout his slide, we have:
101*9.81 - fricforce = 101*3.2 (where gravitational acceleration = 9.81m/s2)
rearrange to get fricforce:
fricforce = 101*9.81 - 101*3.2 = 101*6.61 N
$${\mathtt{fricforce}} = {\mathtt{101}}{\mathtt{\,\times\,}}{\mathtt{6.61}} \Rightarrow {\mathtt{fricforce}} = {\mathtt{667.61}}$$ Newtons
