# friction and sliding box

• Oct 26th 2007, 01:45 PM
rcmango
friction and sliding box
I know there was a similiar problem i did awhile back involving true weight, that was elevator related, but it didn't include friction.

also, i thought when an object is stationary then the Force was 0.

friction is causing me some confusion here.

A 6.40 kg box is sliding across the horizontal floor of an elevator.
The coefficient of kinetic friction between the box and the floor is 0.460.

SO, determine the kinetic frictional force that acts on the box for each of the following cases.

(a) The elevator is stationary.
in Newtons

(b) The elevator is accelerating upward with an acceleration whose magnitude is 1.90 m/s^2.
in Newtons

(c) The elevator is accelerating downward with an acceleration whose magnitude is 1.90 m/s^2.
in Newtons
• Oct 27th 2007, 12:46 AM
CaptainBlack
Quote:

Originally Posted by rcmango
I know there was a similiar problem i did awhile back involving true weight, that was elevator related, but it didn't include friction.

also, i thought when an object is stationary then the Force was 0.

friction is causing me some confusion here.

A 6.40 kg box is sliding across the horizontal floor of an elevator.
The coefficient of kinetic friction between the box and the floor is 0.460.

SO, determine the kinetic frictional force that acts on the box for each of the following cases.

(a) The elevator is stationary.
in Newtons

(b) The elevator is accelerating upward with an acceleration whose magnitude is 1.90 m/s^2.
in Newtons

(c) The elevator is accelerating downward with an acceleration whose magnitude is 1.90 m/s^2.
in Newtons

You do all these in the same way as for a stationary lift, that is the frictional
force is $F=\mu_k m g$, but here $g$ is the effective accelaeration due to gravity,
in the three cases this is $g_0$, $g_0+1.9$ and $g_0-1.9$, where $g_0$ is the conventional
acceleration due to gravity $g_0\approx 9.81 \mbox{m/s}^2$

RonL