dttom Posted November 25, 2006 Posted November 25, 2006 I am having my exam very soon and I have a question want to ask... it is that, friction between two bodies is determined by the equation, f=Nu, where f denotes the friction force; N for the normal force and; u for the friction coefficient, however, it is also known that the resulted kinetic friction force is proportional to the velocity of the sliding object. So does it mean that the velocity of the sliding object would affect the normal force? Or affect the friction coefficient? And I wonder the explanation for this phenomenon.
EvoN1020v Posted November 25, 2006 Posted November 25, 2006 Velocity won't affect the object's normal force because velocity is a vector that is independent from the normal force. You should know that normal force is always perpendicular to the object's centre of mass. The normal force of an object differs depending on the slope of the surface. For instance, an object of 10 kg on a flat surface: 98.1 N An object of 10 kg on a 45 degrees slope: 69.4 N In conclusion, velocity does not affect the object's normal force.
5614 Posted November 25, 2006 Posted November 25, 2006 Well [math]F = \mu R [/math] is an approximation of the frictional forces on a static body. When a body begins to move [math]\mu[/math] changes value (usually a decrease) to [math]\mu_k [/math] (coefficient of kinetic friction). So [math]\mu[/math] changes. We know that [math]F = \mu R [/math] is only approximation. We also know that the frictional force on a moving body varies with its velocity. So I would have as a guess that when the velocity is significantly changing a different, more accurate, formula is required.
Klaynos Posted November 25, 2006 Posted November 25, 2006 [math] F = \mu R [/math] Is the maximum frictional force normally, as 5614 says for a proper friction equation mu is velocity dependent.
EvoN1020v Posted November 25, 2006 Posted November 25, 2006 [math]F = \mu R [/math] Is the maximum frictional force normally, as 5614 says for a proper friction equation mu is velocity dependent. You are implying that air resistance is included in the friction forces. For basic friction with air resistance negligble, velocity is not dependent.
Klaynos Posted November 25, 2006 Posted November 25, 2006 You are implying that air resistance is included in the friction forces. For basic friction with air resistance negligble, velocity is not dependent. No I'm not. Yes it is.
EvoN1020v Posted November 25, 2006 Posted November 25, 2006 I disagree with you because you don't input any velocity values in the formula of [math]F_f=\mu F_N[/math], therefore it's not dependent.
Klaynos Posted November 25, 2006 Posted November 25, 2006 I disagree with you because you don't input any velocity values in the formula of [math]F_f=\mu F_N[/math], therefore it's not dependent. I think you'll find I said the mu ([math]\mu[/math]) was velocity dependent, and that [math]F_f=\mu F_N[/math] was general taken to be the maximum value unless [math]F_f=\mu(v) F_N[/math] where [math]\mu(v)[/math] was defined. Just not quite in those words. Although for most simplifications ffriction is taken to be equal to fapplied when less than [math]\mu F_N[/math]
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