Graph of frictional force vs angle of inclination

[jontyjashan]

Our teacher had asked us to plot graph of frictional force acting on a given mass kept on an incline vs its angle of inclination .i made a graph where the max frictional force was at angle arctan u(coefficient of friction ) . The graph of friction at angle less than arctan u was proportional to sinx(angle of inclination ) and the graph greater than arctan u was proportional to cosx. Am I right,

 

[cepheid]

Our teacher had asked us to plot graph of frictional force acting on a given mass kept on an incline vs its angle of inclination .i made a graph where the max frictional force was at angle arctan u(coefficient of friction ) . The graph of friction at angle less than arctan u was proportional to sinx(angle of inclination ) and the graph greater than arctan u was proportional to cosx. Am I right,

Yeah, that seems right. In the regime where static friction applies and the block remains motionless, friction provides whatever force is required to keep the block from sliding. In other words, the friction force is equal (in magnitude) to the component of the block's weight that is parallel to the incline:

$$F_f = mg\sin\theta$$

where theta is the angle of the incline. But there also is a maximum possible static frictional force:

$$F_f \leq \mu_s F_N = \mu_s mg\cos\theta$$

where F_N is the normal force. Therefore, the largest inclination at which the block won't slide is the inclination at which the required frictional force (left hand side of equation below) is just equal to the max available frictional force (right hand side below):

$$mg\sin(\theta_{\textrm{max}}) = \mu_s mg\cos(\theta_{\textrm{max}})$$

$$\mu_s = \tan(\theta_{\textrm{max}})$$

$$\theta_{\textrm{max}} = \arctan(\mu_s)$$

So I also agree with you there. Above the maximum inclination angle, we are in the regime of kinetic friction, since the block has begun sliding (the friction force is no longer enough to balance the component of the weight parallel to the incline). In this case, the frictional force becomes:

$$F_f = \mu_k F_N = \mu_k mg\cos\theta$$

and now we have cosine-dependence on theta, rather than a sine-dependence. This is significant, because it means that the frictional force actually diminishes with increasing angle (due to the diminishing normal force), which is the opposite trend from before in the static regime, where the higher the angle was, the higher the friction became to prevent sliding (up to a limit). Note also that the constant of proportionality is different here, since you now have a factor of $\mu_k$ in front. You'll probably want to indicate that on your sketch.

 

[jontyjashan]

Thanks. Explanation was really good. Can you make a graph of this using calculators. I was not able to do so. Take any arbitrary values of theta, mass and friction coefficient . It would really help.