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:
[tex]\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}[/tex]
Now, we add the forces in the horizontal direction:
[tex]\Sigma F_h=F_x-F_f[/tex]
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:
[tex]F_x-F_f=ma[/tex]
The x-component of the applied force is determined using the function cosine:
[tex]\cos 50=\frac{F_x}{F}[/tex]
Now, we multiply both sides by "F":
[tex]F\cos 50=F_x[/tex]
Substituting in the horizontal sum of forces:
[tex]F\cos 50-F_f=ma[/tex]
Now, we solve for the friction force. First, we subtract "Fcos50" from both sides:
[tex]-F_f=ma-F\cos 50[/tex]
Now, we multiply both sides by -1:
[tex]F_f=F\cos 50-ma[/tex]
Now, we substitute the values:
[tex]F_f=(195N)\cos 50-(5kg)(2.9\frac{m}{s^2})[/tex]
Solving the operations:
[tex]F_f=110.84N[/tex]
Now, we use the following formula that relates the friction force and the coefficient of friction:
[tex]F_f=\mu N[/tex]
Where:
[tex]\mu=\text{ coefficient of friction}[/tex]
Now, we determine the normal force by adding the vertical forces:
[tex]N-F_y-mg=0[/tex]
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:
[tex]N-F_y=mg[/tex]
Now, we add the y-component of the force:
[tex]N=mg+F_y[/tex]
To determine the y-component of the force we use the function sine:
[tex]\sin 50=\frac{F_y}{F}[/tex]
Now, we multiply both sides by "F":
[tex]F\sin 50=F_y[/tex]
Substituting in the formula for the normal force:
[tex]N=mg+F\sin 50[/tex]
Now, we substitute the values:
[tex]N=(5kg)(9.8\frac{m}{s^2})+(195N)(\sin 50)[/tex]
Solving the operations:
[tex]N=198.38N[/tex]
Now, we substitute in the formula for the friction force:
[tex]F_f=(198.38N)\mu[/tex]
Now, we divide both sides by 198.38:
[tex]\frac{F_f}{198.38N}=\mu[/tex]
Now, we substitute the value of the friction force:
[tex]\frac{110.84N}{198.38N}=\mu[/tex]
Solving the operations:
[tex]0.56=\mu[/tex]
Therefore, the coefficient of friction is 0.56
