Question

What is the force of friction holding a 225 kg box on a ramp that forma a 25 degree angle with the ground?

Answer 1

Answer on Question 48099, Physics, Mechanics | Kinematics | Dynamics |

Question:

What is the force of friction holding a $225\mathrm{kg}$ box on a ramp that forms a 25 degree angle with the ground?

Solution:

Let us write all forces that act on a box:

$$\overrightarrow {m g} + \overrightarrow {N} + \overrightarrow {F _ {f r}} = 0$$

Then projected the forces on axis $x$ and $y$ we have:

$$m \cdot g \cdot \sin \theta - F _ {f r} = 0,$$ $$- m \cdot g \cdot \cos \theta + N = 0.$$

So, we can find $F_{fr}$ :

$$F _ {f r} = m \cdot g \cdot \sin \theta = 2 2 5 k g \cdot 9. 8 \frac {m}{s ^ {2}} \cdot \sin 2 5 {}^ {\circ} = 9 2 6. 1 N$$

Answer:

$$F _ {f r} = 9 2 6. 1 N$$

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