6.2 Friction  (Page 6/12)

A crate on an accelerating truck

A 50.0-kg crate rests on the bed of a truck as shown in [link] . The coefficients of friction between the surfaces are ${\mu }_{\text{k}}=0.300$ and ${\mu }_{\text{s}}=0.400.$ Find the frictional force on the crate when the truck is accelerating forward relative to the ground at (a) 2.00 m/s ² , and (b) 5.00 m/s ² .

Strategy

The forces on the crate are its weight and the normal and frictional forces due to contact with the truck bed. We start by assuming that the crate is not slipping. In this case, the static frictional force $f_{\text{s}}$ acts on the crate. Furthermore, the accelerations of the crate and the truck are equal.

Solution

1. Application of Newton’s second law to the crate, using the reference frame attached to the ground, yields
   
   $\begin{array}{cccccccc}\hfill {\displaystyle \sum F_x}& =\hfill & m{a}_x\hfill & & & \hfill {\displaystyle \sum F_y}& =\hfill & m{a}_y\hfill \\ \hfill f_{\text{s}}& =\hfill & (50.0\text{kg})(2.00{\text{m/s}}^2)\hfill & & & \hfill N-4.90\times {10}^2\text{N}& =\hfill & (50.0\text{kg})(0)\hfill \\ & =\hfill & 1.00\times {10}^2\text{N}\hfill & & & \hfill N& =\hfill & 4.90\times {10}^2\text{N}\text{.}\hfill \end{array}$
   
   We can now check the validity of our no-slip assumption. The maximum value of the force of static friction is
   
   ${\mu }_{\text{s}}N=(0.400)(4.90\times {10}^2\text{N})=196\text{N,}$
   
   whereas the actual force of static friction that acts when the truck accelerates forward at $2.00{\text{m/s}}^2$ is only $1.00\times {10}^2\text{N}\text{.}$ Thus, the assumption of no slipping is valid.
2. If the crate is to move with the truck when it accelerates at $5.0{\text{m/s}}^2,$ the force of static friction must be
   
   $f_{\text{s}}=m{a}_x=(50.0\text{kg})(5.00{\text{m/s}}^2)=250\text{N}\text{.}$
   
   Since this exceeds the maximum of 196 N, the crate must slip. The frictional force is therefore kinetic and is
   
   $f_k={\mu }_{k}N=(0.300)(4.90\times {10}^2\text{N})=147\text{N}\text{.}$
   
   The horizontal acceleration of the crate relative to the ground is now found from
   
   $\begin{array}{ccc}\hfill {\displaystyle \sum F_x}& =\hfill & m{a}_x\hfill \\ \hfill 147\text{N}& =\hfill & (50.0\text{kg})a_x,\hfill \\ \hfill \text{so}{a}_x& =\hfill & 2.94{\text{m/s}}^2.\hfill \end{array}$

Significance

Relative to the ground, the truck is accelerating forward at $5.0{\text{m/s}}^2$ and the crate is accelerating forward at $2.94{\text{m/s}}^2$ . Hence the crate is sliding backward relative to the bed of the truck with an acceleration $2.94{\text{m/s}}^2-5.00{\text{m/s}}^2=\mathrm{-2.06}{\text{m/s}}^2\text{.}$

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