# How do I find the minimum force to be applied to a block sliding in a incline such as it moves with constant velocity?

The problem is as follows:

The figure shows a block over an incline. Find the minimum force that must be applied to the block so that the body of mass $$m=2\,kg$$ such as that body moves with constant velocity upwards in the incline. It is known that the coefficient of friction between the surfaces is $$\mu = 0.3$$ and the angle of the incline is $$\alpha = 30 ^{\circ}$$.

The alternatives given on my book are as follows:

$$\begin{array}{ll} 1.&21\,N\\ 2.&23\,N\\ 3.&18\,N\\ 4.&20\,N\\ 5.&2.2\,N\\ \end{array}$$

I really need help with this problem. Initially I thought that I should decompose the force and weight. Which I assumed that from the figure the force is parallel to the floor which is the base of the incline.

By doing this and considering that the coefficient of friction (which I assumed that is static) this would be translated as follows:

$$F\cos\alpha - \mu N = 0$$

The normal or the reaction from the incline I found it using this logic:

$$N- mg \cos\alpha - F\sin\alpha = 0$$

$$N= mg \cos\alpha + F\sin\alpha$$

Inserting this in the above equation:

$$F\cos\alpha - \mu \left(mg \cos\alpha + F\sin\alpha\right) = 0$$

$$F\cos\alpha - \mu mg \cos\alpha - \mu F\sin\alpha = 0$$

$$F \left( \cos\alpha - \mu \sin\alpha \right) = \mu mg \cos\alpha$$

$$F=\frac{\mu mg \cos\alpha}{\cos\alpha - \mu \sin\alpha}$$

Therefore by inserting there the given information would become into:

$$F=\frac{\frac{3}{10} (2\times 10) \cos 30^{\circ}}{\cos 30^{\circ} - \frac{3}{10} \sin 30^{\circ}}$$

$$F=\frac{\frac{6\sqrt {3}}{2}}{\cos 30^{\circ} - \frac{3}{10} \sin 30^{\circ}}$$

$$F=\frac{\frac{6\sqrt {3}}{2}}{\frac{\sqrt{3}}{2} - \frac{3}{10} \times \frac{1}{2}}$$

Here's where simplification becomes ugly:

$$F=\frac{3\sqrt{3}}{\frac{10\sqrt{3}-3}{20}}$$

$$F=\frac{60\sqrt{3}}{10\sqrt{3}-3} \approx 27.25$$

Therefore in the end I obtain that value for the force. But it is nowhere near to the answers. Can somebody help me with this?. What could I had done wrong?. How could I simplify this?. Can somebody offer a FBD help for this problem?

• What is the force needed for equilibrium? That force is the minimum needed so the block moves upward at a constant velocity of zero feet per second. There is no minimum force that will push the block upward at a constant positive velocity as there is no minimum positive velocity. Nov 18, 2019 at 11:33

Hint.

Calling $$\beta = \frac{\pi}{2}-\alpha$$ and projecting the forces along the incline and assuming the movement positive upwards,

$$F\sin\beta - m g\sin\alpha = \mu\left(F\cos\beta+m g\cos\alpha\right)$$

Free Body Diagram of the mass 'm'

From the vertical (with respect to the slope) forces, we get:
N = mg$$\cos θ$$ + F$$\sin θ$$

So, by balancing the horizontal (with respect to the slope) forces, we get:

mg$$\sin θ$$ + f = F$$\cos θ$$
mg$$\sin θ$$ + μN = F$$\cos θ$$
mg$$\sin θ$$ + μ(mg$$\cos θ$$ + F$$\sin θ$$) = F$$\cos θ$$
mg$$\sin θ$$ + μmg$$\cos θ$$ + μF$$\sin θ$$ = F$$\cos θ$$
mg$$\sin θ$$ + μmg$$\cos θ$$ = F($$\cos θ$$ - μ$$\sin θ$$)
mg($$\sin θ$$ + μ$$\cosθ$$) = F($$\cos θ$$ - $$μ\sin θ$$)

Dividing both sides by $$\cos θ$$, we get:
mg($$\tan θ$$ + μ) = F(1 - μ$$\tan θ$$)
F = mg$$\frac{\tan θ + μ}{1 - μ\tan θ}$$

Now, substituting θ = 30°, m = 2 kg, μ = 0.3 and g = 10 ms2, we get:
F = $$2*10\frac{\frac {1}{\sqrt{3}} + 0.3}{1 - 0.3*\frac{1}{\sqrt{3}}}$$
$$20\frac{0.57 + 0.3}{1 - 0.57*0.3}$$
$$20\frac{0.87}{0.829}$$
$$20*1.04$$
$$20.8$$
~ 21 N