Some Particular Forces — Question 8

Question 8

A prescribed tangential contact force is distinct from the normal

A 2 kg block slides right on a horizontal surface. A 9 N horizontal pull acts right. The surface exerts a measured 3 N resistance left. Take $g=10$ m/s$^2$; vertical acceleration is zero.

1. Find normal force and horizontal acceleration.

2. Find the magnitude and direction of the total surface-contact force.

3. Why must the 3 N not be automatically replaced by the normal force?

Question 8 - Solution

A prescribed tangential contact force is distinct from the normal

(a) Separate perpendicular equations.

$$N-20=0,\quad9-3=2a_x\ \Rightarrow\ \boxed{N=20\,\mathrm N,\quad a_x=3\,\mathrm{m/s^2}}$$ $N=20$ N. Horizontal resultant is $9-3=6$ N, so $a_x=3$ m/s$^2$.

(b) Combine the two contact components.

$$\vec F_{\mathrm{surface}}=(-3,20)\,\mathrm N,\quad F_{\mathrm{surface}}=\sqrt{409}=20.224\,\mathrm N,\quad\phi=\tan^{-1}(3/20)=8.531^\circ$$ The surface force is $(-3,20)$ N, magnitude $\sqrt{409}\approx20.224$ N. Its direction is $\tan^{-1}(3/20)\approx8.531^\circ$ left of upward vertical.

(c) Keep normal and tangential information distinct.

$$\vec F_{\mathrm{surface}}=N\hat y-f\hat x,\quad\hat x\cdot\hat y=0,\quad f=3\,\mathrm N\ne N$$ The normal is perpendicular to the surface; resistance is tangential. Their magnitudes need not agree. No coefficient law is supplied here; the tangential force is measured, with detailed friction models reserved for later sections.