Applying Newton's Laws — Question 3

Question 3

An incline changes the gravitational driving component

Take $g=10$ m/s$^2$. A 4 kg block on a frictionless $30^\circ$ incline connects over an ideal fixed pulley at the top to a hanging 3 kg mass. The light cord is taut.

1. Choose and verify the direction of acceleration.

2. Find acceleration, tension and incline normal force.

3. What hanging mass would instead produce zero acceleration?

Question 3 - Solution

An incline changes the gravitational driving component

(a) Compare driving components.

$$F_{\mathrm{drive}}=m_hg-m_ig\sin30^\circ=30-20=10\,\mathrm N>0$$ The block’s downslope gravity component is $4(10)\sin30^\circ=20$ N. The hanging weight is 30 N, so the 3 kg mass descends and the block ascends.

(b) Write both equations with consistent senses.

$$30-T=3a,\quad T-20=4a:\quad\boxed{a=10/7,\ T=180/7,\ N=20\sqrt3}\quad\text{(SI)}$$ $30-T=3a$ and $T-20=4a$ give $a=10/7\approx1.429$ m/s$^2$ and $T=180/7\approx25.714$ N. Perpendicular balance gives $N=20\sqrt3\approx34.641$ N.

(c) Match the driving weights.

$$a=0\ \Rightarrow\ m_hg=m_ig\sin30^\circ\ \Rightarrow\ m_h=2\,\mathrm{kg}$$ Zero acceleration requires $m_hg=4g\sin30^\circ$, hence $m_h=2$ kg. This permits either rest or compatible constant motion, depending on initial conditions.