Question

a 440 kg bear gradping vertical trees slides down at constant velocity. what is the friction force that acts on the bear

Answer 1

If the bear is moving at a CONSTANT velocity, that means that the bear is NOT accelerating.

If there's no acceleration, there cannot be any force:

F = m*a

but if a = 0, then F = 0 as well.

There are two forces acting on the bear as he slides down:

1) Force of gravity F(grav) acting downward

2) Force of friction F(fric) acting upward

The TOTAL force we determined is zero, and is also the SUM of both of the two forces:

F = total force = F(fric) + F(grav) = 0

Using the above equation we find:

F(fric) = -F(grav)

We can easily find F(grav):

F(grav) = mass x gravity = m*g

We know m = mass of bear = 450 kgand g = -9.81 m/s^2, so:

F(grav) = m*g = 500kg * -9.81 m/s^2 = -4414.5 N

(the "-" is because gravity acts DOWN)

Now we can solve for the force of friction:

F(fric) = -F(grav) = - (-4414.5 N) = 4414.6 N

So the force of friction is 4905 N (acting up the tree).