. Thus, we know that the normal force acting on the box will equal its weight. If friction is adequate, we expect that the box will accelerate in the x-direction at the same rate as the truck does. In that case, we expect: | Latex |
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\begin{large}\[ F_{f} = (15\:{\rm kg})(3.5\:{\rm m/s}^{2}) = 53\:{\rm N} \]\end{large} |
| Warning |
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We're not finished yet! |
It is important now to check that this result does not conflict with the requirement that | Latex |
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\begin{large} \[ F_{f} \le \mu_{s} N\] \end{large} |
Since we have already used the y-direction equation of Newton's Second Law to conclude that the normal force on the box is equal to mg, we find: | Latex |
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\begin{large} \[ F_{f} = 53\:{\rm N} < \mu_{s} N = 0.40(15\:{\rm kg})(9.8\:{\rm m/s}^{2}) = 59\:{\rm N} \]\end{large} |
We therefore conclude that our answer, Ff = 53 N, is compatible with the static friction limit. | 