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The box shown on the rough ramp above is sliding up the ramp.calculate the force of friction on the box

The box shown on the rough ramp above is sliding up the ramp.calculate the force of-example-1

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We are given a box that slides up a ramp. To determine the force of friction we will use the following relationship:


F_f=\mu N

Where.


\begin{gathered} N=\text{ normal force} \\ \mu=\text{ coefficient of friction} \end{gathered}

To determine the Normal force we will add the forces in the direction perpendicular to the ramp, we will call this direction the y-direction as shown in the following diagram:

In the diagram we have:


\begin{gathered} m=\text{ mass}_{} \\ g=\text{ acceleration of gravity} \\ mg=\text{ weight} \\ mg_y=y-\text{component of the weight. } \end{gathered}

Adding the forces in the y-direction we get:


\Sigma F_y=N-mg_y

Since there is no movement in the y-direction the sum of forces must be equal to zero:


N-mg_y=0

Now we solve for the normal force:


N=mg_y

To determine the y-component of the weight we will use the trigonometric function cosine:


\cos 40=(mg_y)/(mg)

Now we multiply both sides by "mg":


mg\cos 40=mg_y

Now we substitute this value in the expression for the normal force:


N=mg\cos 40

Now we substitute this in the expression for the friction force:


F_f=\mu mg\cos 40

Now we substitute the given values:


F_f=(0.2)(10\operatorname{kg})(9.8(m)/(s^2))\cos 40

Solving the operations:


F_f=15.01N

Therefore, the force of friction is 15.01 Newtons.

The box shown on the rough ramp above is sliding up the ramp.calculate the force of-example-1
User Kateryna
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