Question

There is an unknown container on a horizontal surface that you need to

move. The coefficient of friction for the floor is 0.54 and the box

weighs 250kg. If you apply a force of 350N at an angle of 50 degrees

the box will accelerate at a rate what rate?

Answer 1

Answer on Question 57046, Physics, Mechanics, Relativity

Question:

There is an unknown container on a horizontal surface that you need to move. The coefficient of friction for the floor is 0.54 and the box weighs $250kg$ . If you apply a force of $350N$ at an angle of 50 degrees the box will accelerate at a rate what rate?

Solution:

Let's draw a free-body diagram and write all forces that act on a container:

$$m \vec {g} + \vec {N} + \overrightarrow {F _ {f r}} + \vec {F} = m \vec {a}$$

Then projected the forces on axis $x$ we get $(F_{x} = F\cos \alpha)$ :

$$F \cos \alpha - F _ {f r} = m a, (1)$$

And projected the forces on axis $y$ we get $(F_y = F\sin \alpha)$ :

$$N + F \sin \alpha - m g = 0,$$ $$N = m g - F \sin \alpha .$$

Then we can find the friction force:

$$F _ {f r} = \mu N = \mu (m g - F \sin \alpha)$$

Substituting $F_{fr}$ into the equation (1) we can find the acceleration of the container:

$$F \cos \alpha - \mu (m g - F \sin \alpha) = m a,$$ $$a = \frac {F \cos \alpha - \mu (m g - F \sin \alpha)}{m}. (2)$$

Unfortunately, under such conditions of the question, an unknown container will not move, because the applied force less than the friction force needed to move object:

$$F _ {x} = F \cos \alpha = 350N \cdot \cos 50{}^\circ = 225N.$$ $$F _ {fr} = \mu N = \mu (mg - F \sin \alpha) = 0.54 \cdot \left(250kg \cdot 9.8 \frac{m}{s^2} - 350N \cdot \sin 50{}^\circ\right) = 1178N.$$ $$F _ {x} < F _ {fr}.$$

So, maybe you mistake when enter the data for this question. If you substitute the correct data into the equation (2), you find the acceleration of an unknown container.

Answer:

$$a = \frac {F \cos \alpha - \mu (mg - F \sin \alpha)}{m}.$$

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