Some Particular Forces — Question 4

Question 4

An angled pull can reduce floor support

A 5 kg block stays on a frictionless horizontal floor. A force of 20 N acts $30^\circ$ above horizontal. Take $g=10$ m/s$^2$.

1. Find the normal force.

2. Find horizontal acceleration.

3. At what pull magnitude would contact become marginal if the angle stayed $30^\circ$?

Question 4 - Solution

An angled pull can reduce floor support

(a) Balance the constrained vertical component.

$$N+P\sin\theta-mg=0\ \Rightarrow\ \boxed{N=50-20\sin30^\circ=40\,\mathrm N}$$ $N+20\sin30^\circ-50=0$, so $N=40$ N.

(b) Use the unbalanced horizontal component.

$$ma_x=P\cos\theta\ \Rightarrow\ a_x=\frac{20\cos30^\circ}{5}=2\sqrt3\,\mathrm{m/s^2}$$ $a_x=20\cos30^\circ/5=2\sqrt3\approx3.464$ m/s$^2$.

(c) Set the limiting normal to zero.

$$N=0\ \Rightarrow\ P_{\mathrm{crit}}=\frac{mg}{\sin\theta}=\frac{50}{0.5}=100\,\mathrm N$$ $P\sin30^\circ=50$ gives $P=100$ N. Above that value the assumed zero vertical acceleration with contact would require negative N; the block lifts and the constrained equations must change.