0
$\begingroup$

What forces are acting on a box placed on the loading platform of a truck that is accelerating? How does the truck's acceleration affect the box, and why does the box not slide back and forth if the acceleration is not too high? How does friction come into play in this scenario?

$\endgroup$
1
  • $\begingroup$ What are your own thoughts on these questions $\endgroup$
    – Bob D
    Jun 6, 2023 at 20:56

1 Answer 1

1
$\begingroup$

What forces are acting on a box placed on the loading platform of a truck that is accelerating?

It depends on the magnitude of the acceleration, the weight of the box, and the coefficients of static and kinetic friction.

How does the truck's acceleration affect the box...

It determines whether or not the box slides on the truck.

...and why does the box not slide back and forth if the acceleration is not too high?

If the acceleration is not too high, meaning the force acting on the box by the truck does not exceed the maximum possible static friction force, then static friction will prevent the box from sliding back and forth. The maximum possible static friction force is $\mu_{s}mg$ where $\mu$ is the coefficient of static friction and $m$ is the mass of the box.

How does friction come into play in this scenario?

The static friction is a variable force determines whether or not the box slides on the truck. The static friction force equals and opposes the applied force up until the maximum possible static friction force is exceeded. If it is not exceeded, the box will not slide. If it is exceeded, then the box will slide as friction becomes kinetic (sliding friction), which determines the acceleration (if any) of the box relative to the truck. The kinetic friction force, which is generally assume constant, is $\mu_{k}mg$ where $\mu_{k}$ is the coefficient of kinetic friction. Note that, in general, $\mu_{k}\lt \mu_s$.

For more information, I recommend you visit the following site: http://hyperphysics.phy-astr.gsu.edu/hbase/frict2.html#kin

Hope this helps.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.