"Weightlessness" is analogous to free-fall (neglecting air resistance), during which the only force on an object is the force of gravity. When normal force becomes zero, the object loses physical contact with the surface.

When normal force is equal to the force of gravity, the object is in equilibrium and the net vertical force is zero. This results in zero acceleration, but does not result in "weightlessness." Mass is not a force, and negative force values merely indicate relative direction (assuming correct calculations)

Frictional force is calculated using the normal force and the friction coefficient. Since the book is at rest (not in motion), we use the coefficient of static friction.

\(\displaystyle F_f=\mu_sF_N\)

The normal force will be equal and opposite the force pushing directly into the surface; in this case, that means the normal force will be 20N.

Solve for the force of friction:

\(\displaystyle F_f=(0.3)(20N)=6N\)

\(\displaystyle \begin{align*}&\text{To solve a problem of force balance, a good idea is to draw}\\&\text{it out. Since the block is at rest, i.e. stationary, the sum}\\&\text{of forces acting on it must be equal to zero. The forces acting}\\&\text{on it are gravity (in the form of the mass times the gravitational}\\&\text{constant), the force of the rope supporting it, a normal force,}\\&\text{and since it’s on an incline, a frictional force to balance}\\&\text{it all out. Notice that the tension in the rope mitigates some}\\&\text{of the gravitational force (as illustrated in the upper left}\\&\text{corner) to lessen what is applied to the slope:}\\&F_{mass}=m*g\\&F_{normal}=(F_{mass}-F_{rope})*cos(a)\\&F_{friction}=(F_{mass}-F_{rope})*sin(a)\\&F_{normal}=(190kg(9.81\frac{m}{s^2})-559.17N)(cos(33^{\circ}))\\&F_{normal}=1094.24N\end{align*}\)

\(\displaystyle \begin{align*}&\text{To solve a problem of force balance, a good idea is to draw}\\&\text{it out. Since the block is at rest, i.e. stationary, the sum}\\&\text{of forces acting on it must be equal to zero. The forces acting}\\&\text{on it are gravity (in the form of the mass times the gravitational}\\&\text{constant), a normal force, and since it’s on an incline, a frictional}\\&\text{force to balance it all out:}\\&F_{mass}=m*g\\&F_{normal}=m*g*cos(a)\\&F_{friction}=m*g*sin(a)\\&F_{normal}=233kg(9.81\frac{m}{s^2})(cos(81^{\circ}))\\&F_{normal}=357.57N\end{align*}\)

\(\displaystyle \begin{align*}&\text{To solve a problem of force balance, a good idea is to draw}\\&\text{it out. Since the block is at rest, i.e. stationary, the sum}\\&\text{of forces acting on it must be equal to zero. The forces acting}\\&\text{on it are gravity (in the form of the mass times the gravitational}\\&\text{constant), a normal force, and since it’s on an incline, a frictional}\\&\text{force to balance it all out:}\\&F_{mass}=m*g\\&F_{normal}=m*g*cos(a)\\&F_{friction}=m*g*sin(a)\\&F_{normal}=187kg(9.81\frac{m}{s^2})(cos(83^{\circ}))\\&F_{normal}=223.57N\end{align*}\)

\(\displaystyle \begin{align*}&\text{To solve a problem of force balance, a good idea is to draw}\\&\text{it out. Since the block is at rest, i.e. stationary, the sum}\\&\text{of forces acting on it must be equal to zero. The forces acting}\\&\text{on it are gravity (in the form of the mass times the gravitational}\\&\text{constant), a normal force, and since it’s on an incline, a frictional}\\&\text{force to balance it all out:}\\&F_{mass}=m*g\\&F_{normal}=m*g*cos(a)\\&F_{friction}=m*g*sin(a)\\&F_{normal}=267kg(9.81\frac{m}{s^2})(cos(49^{\circ}))\\&F_{normal}=1718.4N\end{align*}\)

\(\displaystyle \begin{align*}&\text{To solve a problem of force balance, a good idea is to draw}\\&\text{it out. Since the block is at rest, i.e. stationary, the sum}\\&\text{of forces acting on it must be equal to zero. The forces acting}\\&\text{on it are gravity (in the form of the mass times the gravitational}\\&\text{constant), the force of the rope supporting it, a normal force,}\\&\text{and since it’s on an incline, a frictional force to balance}\\&\text{it all out. Notice that the tension in the rope mitigates some}\\&\text{of the gravitational force (as illustrated in the upper left}\\&\text{corner) to lessen what is applied to the slope:}\\&F_{mass}=m*g\\&F_{normal}=(F_{mass}-F_{rope})*cos(a)\\&F_{rope}=F_{mass}-\frac{F_{normal}}{cos(a)}\\&F_{rope}=1098.72N-\frac{417.98N}{cos(18^{\circ})}\\&F_{rope}=659.23N\end{align*}\)