Answer:
[tex]\displaystyle T=48.86\ N[/tex]
Explanation:
Net Force
The second Newton's law explains how to understand the dynamics of a system where several forces are acting. The forces are vectorial magnitudes which means the x and y coordinates must be treated separately. For each component, the net force must equal the mass by the acceleration, i.e.
[tex]F_{nx}=ma_x[/tex]
[tex]F_{ny}=ma_y[/tex]
The box with mass m=20 kg is pulled by a rope with a [tex]\theta= 30^o[/tex] angle above the horizontal. It means that force (called T) has two components:
[tex]T_x=Tcos\theta[/tex]
[tex]T_y=Tsin\theta[/tex]
We'll assume the positive directions are to the right and upwards and that the box is being pulled to the right. There are two forces in the x-axis: The x-component of T (to the right) and the friction force (to the left). So the equilibrium equation for x is
[tex]\displaystyle T\ cos\theta -Fr=m.a[/tex]
There are three forces acting in the y-axis: The component of T (upwards), the weight (downwards), and the Normal (upwards). Since there is no movement in the y-axis, the net force is zero and:
[tex]\displaystyle N+T\ sin\theta -mg=0[/tex]
Rearranging:
[tex]\displaystyle N+T\ sin\theta =mg[/tex]
Solving for N in the y-axis:
[tex]\displaystyle N=mg-T\ sin\theta[/tex]
The friction force is given by
[tex]\displaystyle Fr=\mu.N[/tex]
Replacing in the equation for the x-axis, we have
[tex]\displaystyle T\ cos\theta -\mu\ N=ma[/tex]
Replacing the formula for N in the equation for the x-axis
[tex]\displaystyle T\ cos\theta -\mu(mg-T\ sin\theta)=ma[/tex]
Operating and rearranging
[tex]\displaystyle T\ cos\theta -\mu\ mg+T\ \mu\ sin\theta=ma[/tex]
[tex]\displaystyle T\ (cos\theta +\mu\ sin\theta)=ma +\mu\ mg[/tex]
Solving for T:
[tex]\displaystyle T=\frac{a+\mu\ g}{cos\theta +\mu\ sin\theta }\ m[/tex]
Plugging in the given values:
[tex]\displaystyle T=\frac{0.4+0.2(9.8)}{cos30^o+0.2\ sin30^o }\ .20[/tex]
[tex]\boxed{\displaystyle T=48.86\ N}[/tex]
