Question

Figure shows a block of mass m resting on a 20∘ slope. The block has coefficients of friction 0.81 and 0.45 with the surface. It is connected via a massless string over a massless, frictionless pulley to a hanging block of mass 2.0 kg.
If this minimum mass is nudged ever so slightly, it will start being pulled up the incline. What acceleration will it have?

Answer 1

Answer on Question #48063-Engineering-SolidWorks-CosmoWorks-Ansys

Figure shows a block of mass $m$ resting on a $20^{\circ}$ slope. The block has coefficients of friction $\mu_{s} = 0.81$ and $\mu_{k} = 0.45$ with the surface. It is connected via a massless string over a massless, frictionless pulley to a hanging block of mass $m_{2} = 2.0 \, kg$.

If this minimum mass is nudged ever so slightly, it will start being pulled up the incline. What acceleration will it have?

Solution

On a $20^{\circ}$ slope, the force parallel causes acceleration down the incline

$$F _ {p} = m \cdot g \cdot \sin \theta = m \cdot 9. 8 \cdot \sin 2 0 ^ {\circ}.$$

The friction force is caused by the force perpendicular $m \cdot g \cdot \cos \theta$ is

$$F _ {f r} = \mu_ {s} m \cdot g \cdot \cos \theta = 0. 8 1 \cdot m \cdot 9. 8 \cdot \cos 2 0 ^ {\circ}.$$

Since you are trying to move the block up the slope, both of these forces are opposing motion:

$$F _ {p} + F _ {f r} = m \cdot 9. 8 \cdot \sin 2 0 ^ {\circ} + 0. 8 1 \cdot m \cdot 9. 8 \cdot \cos 2 0 ^ {\circ} = m \cdot 9. 8 (\sin 2 0 ^ {\circ} + 0. 8 1 \cdot \cos 2 0 ^ {\circ}).$$

The force caused by the $2\mathrm{kg}$ is $F = 2\cdot 9.8 = 19.6N$ pulling the block up the slope.

The block will stick and not slip, if the sum of the down forces is less than 19.6 N.

$$m \cdot 9. 8 (\sin 2 0 ^ {\circ} + 0. 8 1 \cdot \cos 2 0 ^ {\circ}) < 1 9. 6.$$ $$\mathrm {m} < \frac {1 9 . 6}{9 . 8 (\sin 2 0 ^ {\circ} + 0 . 8 1 \cdot \cos 2 0 ^ {\circ})}.$$ $$\mathrm {m} < 1. 8 1 \mathrm {k g}.$$

If this minimum mass is nudged ever so slightly, it will start being pulled up the incline. What acceleration will it have?

We need this kinetic friction force:

$$F _ {k} = \frac {\mu_ {k}}{\mu_ {s}} F _ {f r} = \frac {\mu_ {k}}{\mu_ {s}} \mu_ {s} m \cdot g \cdot \cos \theta = \mu_ {k} m \cdot g \cdot \cos \theta = 0. 4 5 \cdot 1. 8 1 \cdot 9. 8 \cdot \cos 2 0 ^ {\circ}.$$

So,

$$1 9. 6 - 1. 8 1 \cdot 9. 8 (\sin 2 0 ^ {\circ} + 0. 4 5 \cdot \cos 2 0 ^ {\circ}) = 1. 8 1 \cdot a.$$ $$a = 3. 3 \frac {m}{s ^ {2}}.$$

Answer: $3.3 \frac{m}{s^2}$.

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