Question

At the instant shown a 195 N force accelerates a 5 kg block at a rate of 2.9 m/s/s. Determine the coefficient of friction if = 50 degrees.

Answer 1

We are asked to determine the coefficient of friction of the 5 kg block. The free-body diagram of the system is the following:

Where:

`\begin{gathered} F=\text{ applied force} \\ F_x=\text{ x-component of applied force} \\ F_y=\text{ y-component of the applied force} \\ F_f=\text{ friction force} \\ m=\text{ mass} \\ g=\text{ acceleration of gravity} \\ N=\text{ normal force} \end{gathered}`

Now, we add the forces in the horizontal direction:

`\Sigma F_h=F_x-F_f`

According to Newton's second law we have that the sum of forces is equal to the product of the mass and the acceleration, therefore, we have:

`F_x-F_f=ma`

The x-component of the applied force is determined using the function cosine:

`\cos 50=(F_x)/(F)`

Now, we multiply both sides by "F":

`F\cos 50=F_x`

Substituting in the horizontal sum of forces:

`F\cos 50-F_f=ma`

Now, we solve for the friction force. First, we subtract "Fcos50" from both sides:

`-F_f=ma-F\cos 50`

Now, we multiply both sides by -1:

`F_f=F\cos 50-ma`

Now, we substitute the values:

`F_f=(195N)\cos 50-(5kg)(2.9(m)/(s^2))`

Solving the operations:

`F_f=110.84N`

Now, we use the following formula that relates the friction force and the coefficient of friction:

`F_f=\mu N`

Where:

`\mu=\text{ coefficient of friction}`

Now, we determine the normal force by adding the vertical forces:

`N-F_y-mg=0`

the sum of forces is equal to zero since there is no acceleration in the vertical direction. Now, we solve for the normal force by adding "mg" to both sides:

`N-F_y=mg`

Now, we add the y-component of the force:

`N=mg+F_y`

To determine the y-component of the force we use the function sine:

`\sin 50=(F_y)/(F)`

Now, we multiply both sides by "F":

`F\sin 50=F_y`

Substituting in the formula for the normal force:

`N=mg+F\sin 50`

Now, we substitute the values:

`N=(5kg)(9.8(m)/(s^2))+(195N)(\sin 50)`

Solving the operations:

`N=198.38N`

Now, we substitute in the formula for the friction force:

`F_f=(198.38N)\mu`

Now, we divide both sides by 198.38:

`(F_f)/(198.38N)=\mu`

Now, we substitute the value of the friction force:

`(110.84N)/(198.38N)=\mu`

Solving the operations:

`0.56=\mu`

Therefore, the coefficient of friction is 0.56