– Sphere mass: $m_s=3mathrm{~kg}$
– Block mass: $m_b=1.2mathrm{~kg}$

**Objective**

– Find the readings on the scale.

The readings on the scale represent the force that is acting on the scale, produced by the objects hanging on it. We can find these readings by considering the Newton’s second law.

Newton’s second law (In one dimension) states:

$$begin{align}
mcdot a=F_1+F_2+F_3+…
end{align}$$

Where $m$ is the mass of the object, $a$ is the acceleration of the object and $F_{1,2,3,…}$ are the forces acting on that object. Since our objects are not moving (they are in equilibrium), they are not accelerating so: $a=0$.

Only two forces are acting on these two objects, one is due to gravity $F_g=mcdot g$ and the other one is tension force due to hanging $T$. Gravity force $F_g$ is acting on the object downwards and tension for is acting on it upwards, so they need to have different sign.

We used the minus sign in front of $F_g$ since it’s pointing downwards. We could have also used $+$ in front of $F_g$ and $-$ in front of $T$, it wouldn’t change the result, as that only changes our coordinate system (What we choose to be positive).

Now, using formula $(4)$, we get:

$$begin{align}
T=F_g=mcdot g
end{align}$$

Now using this formula, we can simply calculate the readings.

Since the sphere is the only object hanging on the lower scale, we can simply use the mass of sphere $m$ in $(5)$, and get the reading $T_{lower}$:

$$begin{aligned}
T_{lower}&=m_scdot g
\&=3cdot 9.81
\&=boxed{29.43mathrm{~N}}
end{aligned}$$

The upper scale has two masses hanging on it: Mass of sphere $m_s$and mass of the block $m_b$ (Mass of lower scale is neglected). So both masses are going to contribute to gravitational force $F_g$:

$$F_g=(m_1+m_2)cdot g$$

Applying this to the formula $(5)$, we can find the reading on the upper scale $T_{upper}$ as:

$$begin{aligned}
T_{upper}&=(m_1+m_2)cdot g
\&=(1.2+3)cdot 9.81
\&=boxed{41.2mathrm{~N}}
end{aligned}$$

$T_{upper}=41.2mathrm{~N}$

$$
begin{equation}
F_{net} = F_{scale} – F_{g} = ma
end{equation}
$$

Since the elevator’s speed is constant, we know that acceleration $vec{a} = 0$ and $F_{net} =0$. Therefore, using Equation 1,

$$
begin{align*}
F_{scale} &= F_g\
&= mg\
&= (75.0 , mathrm{kg})(9.80,mathrm{frac{m}{s^2}})\
&= boxed{735 ,mathrm{N}}
end{align*}
$$

The elevator is slowing down while moving upward with an acceleration $vec{a} = -2.0 ,mathrm{frac{m}{s}}^2$. Therefore, using Equation 1,

$$
begin{align*}
F_{scale} &= F_{net}+F_g\
&= ma+mg \
&= m(a+g)\
&= (75.0,mathrm{kg})(-2.00,mathrm{frac{m}{s^2}} + 9.80,mathrm{frac{m}{s^2}})\
&= boxed{585 ,mathrm{N}}
end{align*}
$$

The elevator is speeding up while moving downward with an acceleration $vec{a} = -2.0,mathrm{frac{m}{s}}^2$. Thus, using Equation 1,

$$
begin{align*}
F_{scale} &= F_{net}+F_g\
&= ma+mg \
&= m(a+g)\
&= (75.0,mathrm{kg})(-2.00,mathrm{frac{m}{s^2}} + 9.80,mathrm{frac{m}{s^2}})\
&= boxed{585 ,mathrm{N}}
end{align*}
$$

The elevator is moving downward at a constant speed, therefor $vec{a} = 0$ and $F_{net} = 0$. Using Equation 1,

$$
begin{align*}
F_{scale} &= F_g\
&= mg\
&= (75.0 , mathrm{kg})(9.80,mathrm{frac{m}{s^2}})\
&= boxed{735 ,mathrm{N}}
end{align*}
$$

The elevator is slowing down to a stop at a constant magnitude of acceleration $vec{a}$. Thus, using Equation 1,

$$
begin{align*}
F_{scale} &= F_{net}+F_g\
&= ma+mg\
&= (75.0,mathrm{kg})(a) + (75.0,mathrm{kg})(9.80,mathrm{frac{m}{s^2}})\
&= boxed{75.0a + 735 ,mathrm{N}}
end{align*}
$$

$textbf{b.)$F_{scale} = 585 ,mathrm{N}$}$

$textbf{c.)$F_{scale} = 585 ,mathrm{N}$}$

$textbf{d.)$F_{scale} = 735 ,mathrm{N}$}$

$$
textbf{e.)$F_{scale} = 75.0a + 735 ,mathrm{N}$}
$$