Question

Please help me with the formular of this question and the solvings.​

Answer 1

Answer:

$\displaystyle T=48.86\ N$

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.

$F_{nx}=ma_x$

$F_{ny}=ma_y$

The box with mass m=20 kg is pulled by a rope with a $\theta= 30^o$ angle above the horizontal. It means that force (called T) has two components:

$T_x=Tcos\theta$

$T_y=Tsin\theta$

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

$\displaystyle T\ cos\theta -Fr=m.a$

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:

$\displaystyle N+T\ sin\theta -mg=0$

Rearranging:

$\displaystyle N+T\ sin\theta =mg$

Solving for N in the y-axis:

$\displaystyle N=mg-T\ sin\theta$

The friction force is given by

$\displaystyle Fr=\mu.N$

Replacing in the equation for the x-axis, we have

$\displaystyle T\ cos\theta -\mu\ N=ma$

Replacing the formula for N in the equation for the x-axis  

$\displaystyle T\ cos\theta -\mu(mg-T\ sin\theta)=ma$

Operating and rearranging

$\displaystyle T\ cos\theta -\mu\ mg+T\ \mu\ sin\theta=ma$

$\displaystyle T\ (cos\theta +\mu\ sin\theta)=ma +\mu\ mg$

Solving for T:

$\displaystyle T=\frac{a+\mu\ g}{cos\theta +\mu\ sin\theta }\ m$

Plugging in the given values:

$\displaystyle T=\frac{0.4+0.2(9.8)}{cos30^o+0.2\ sin30^o }\ .20$

$\boxed{\displaystyle T=48.86\ N}$