# Speed, weight and force [closed]

If a road roller of weight $W$ is rolling on the road at the speed of $X\,\mathrm{kmph}$, how much force does it apply on the surface of the road (considering a sampled surface of the same size as the contact point between roller and road)?

How does the speed of the roller relate to the force that is applied?

Assume it is Earth's surface and Earth's natural gravitational force, and that the surface is flat and the road roller is in constant speed. Also assume friction $Fr$.

Edit:

According to Newton's law: $F = m a$.

If for a point in time $t$ we consider the acceleration $a$ then $a=V / t$.

So

$$F = \frac{m V}{t} + \text{gravitational force} \Leftrightarrow \\ F = \frac{m V}{t} + m g \Leftrightarrow \\ F = \frac{m Vmg}{t}$$

When the roller is in motion, can the motion vector $F$ and the vector $F$ for gravity be summed directly, even though their directions are not the same? Could someone throw some light in this?

• Speed is given as X, a constant, so no acceleration. No information is given on coefficient of friction or size of contact patch. Sometimes questions include superfluous information to get us off track - I believe this is one such question and that the answer is surprisingly simple. Apr 21, 2015 at 6:41
• @IanF1 sure assume friction is Fr Apr 21, 2015 at 6:48
• The point is, friction is horizontal. The force between the road and the roller is vertical. As the roller is not accelerating vertically, the force between the roller and the road must match the weight. F=W (or F=mg) where m is the mass as opposed to the weight). Apr 21, 2015 at 6:50
• Acceleration is rate of change of velocity. Velocity of the roller is not changing so acceleration is zero. Apr 21, 2015 at 6:54
• Yes - as i said, Weight = mass x gravity or W=mg Apr 21, 2015 at 6:58

The only forces acting in the vertical direction are the weight, $W=mg$, and the reaction force between the roller and the road, $F$. As the roller is not accelerating vertically, these must balance:
$$F = W = mg$$