Physics

Dynamics Problems

Solving dynamics problems. Force diagrams, inclined planes, calculation of accelerations, and normal force in elevators and pulleys.

To solve a general dynamics problem we will follow these steps:
1. Free-body diagram:
We draw the forces acting on each body of the system following these rules:
  • Each mass has a weight W=mg
  • On each surface that supports a mass (ground, another body pushing... etc.) there is a Normal force, N
  • If there is a taut string, a tension, T appears.
  • On surfaces where a body tends to slide over another, or on the ground, there is a friction force, FfF_f
2. For each mass we apply Newton's 2nd Law. If there are forces in several axes we will apply the 2nd Law for each axis, decomposing the forces into the chosen axes:
{∑Fx=m1⋅a1x∑Fy=m1⋅a1y{∑Fx=m2⋅a2x∑Fy=m2⋅a2y…\begin{cases} \sum F_x = m_1 \cdot a_{1x} \\ \sum F_y = m_1 \cdot a_{1y} \end{cases} \quad \begin{cases} \sum F_x = m_2 \cdot a_{2x} \\ \sum F_y = m_2 \cdot a_{2y} \end{cases} \quad \dots
We must respect the sign convention when adding forces and accelerations
Sign convention
3. We obtain a system with enough equations. By solving it we obtain the requested result
Calculate the acceleration of the system, knowing that the friction on the plane is μ=0.2\mu=0.2, and the body has a mass of m=15kg
Inclined plane problem statement
1. Free-body diagram of each body:
We draw the isolated body and the internal and external forces acting on it. We take the inclined plane (X-axis) and its perpendicular (Y-axis) as coordinate axes.
Inclined plane free body diagram
We decompose the force W into its components by trigonometry:
Wx=W⋅sin⁡30∘Wy=W⋅cos⁡30∘\begin{aligned} W_x &= W \cdot \sin 30^\circ \\ W_y &= W \cdot \cos 30^\circ \end{aligned}
2. We set up the equations of the system with Newton's 2nd Law. One equation for each axis
{∑Fx=m⋅ax∑Fy=m⋅ay⇒{Wx−Ff=m⋅axN−Wy=m⋅ay\begin{cases} \sum F_x = m \cdot a_x \\ \sum F_y = m \cdot a_y \end{cases} \Rightarrow \begin{cases} W_x - F_f = m \cdot a_x \\ N - W_y = m \cdot a_y \end{cases}
Since there is no motion in Y, ay=0a_y = 0. The system becomes
{W⋅sin⁡30∘−μ⋅N=m⋅axN−W⋅cos⁡30∘=0\begin{cases} W \cdot \sin 30^\circ - \mu \cdot N = m \cdot a_x \\ N - W \cdot \cos 30^\circ = 0 \end{cases}
3. We apply the data from the problem and solve the system:
{73.5−0.2N=15axN−127.3=0⇒a=3.2 m/s2\begin{cases} 73.5 - 0.2N = 15a_x \\ N - 127.3 = 0 \end{cases} \Rightarrow \htmlClass{text-primary}{a = 3.2 \text{ m/s}^2}
A 70 kg person steps on a scale inside an elevator. Calculate the weight the scale will show:
Elevator problem statement
a) When the elevator accelerates upwards at a=2 m/s2a=2\text{ m/s}^2 and
b) When the elevator accelerates downwards at a=2 m/s2a=2\text{ m/s}^2
a) Approach going up:
1. Free-body diagram:
Elevator moving up diagram
2. Newton's 2nd Law equation:
N−W=m⋅aN - W = m \cdot a
The normal force is the reaction force that the scale exerts on the person, which will be the weight it shows, therefore:
N=m(g+a)N = m(g + a)
3. Applying the data we get: N=826 NN = 826\text{ N}
b) Approach going down:
1. Free-body diagram: It is the same as in the previous case but the acceleration is downwards, which will be negative, a=−2 m/s2a=-2\text{ m/s}^2
Elevator moving down diagram
2. Newton's 2nd Law equation: same as the previous one
N=m(g+a)N = m(g + a)
3. Applying the data we get: N=546 NN = 546\text{ N}
A body in dynamic equilibrium can have uniform motion, but it does not experience acceleration, which implies that the net force acting on it is 0. This is the 1st condition for equilibrium
∑F=0\sum \mathbf{F} = 0
Neither can it rotate, since its angular acceleration is 0, which implies the 2nd condition for equilibrium:
∑τ=0\sum \tau = 0
In equilibrium problems, it will be necessary to apply one or both of these equilibrium conditions.
Rope equilibrium statement
The body of mass m=20 kg in the figure is in equilibrium. Calculate the tensions of the ropes
1. Free-body diagram at the knot:
Rope equilibrium diagram
2. Equilibrium equations
∑F=0\sum \mathbf{F} = 0
{∑Fx=0∑Fy=0⇒{T2x−T1=0T2y−W=0\begin{cases} \sum F_x = 0 \\ \sum F_y = 0 \end{cases} \Rightarrow \begin{cases} T_{2x} - T_1 = 0 \\ T_{2y} - W = 0 \end{cases}
Which can be written as:
{T2⋅cos⁡60∘=T1T2⋅sin⁡60∘=m⋅g\begin{cases} T_2 \cdot \cos 60^\circ = T_1 \\ T_2 \cdot \sin 60^\circ = m \cdot g \end{cases}
3. Applying the data we get:
T2=m⋅gsin⁡60∘=226 NT1=T2⋅cos⁡60∘=113 N\begin{aligned} T_2 &= \frac{m \cdot g}{\sin 60^\circ} = 226 \text{ N} \\ T_1 &= T_2 \cdot \cos 60^\circ = 113 \text{ N} \end{aligned}
Calculate the force and its point of application on the 3m bar to lift the weights in the figure in equilibrium:
Bar torque statement
1. Free-body diagram on the bar:
Bar torque diagram
2. Equilibrium equations: we take the left end as the pivot point
∑F=0F−W1−W2=0\begin{aligned} \sum \mathbf{F} &= 0 \\ F - W_1 - W_2 &= 0 \end{aligned}
∑τ=0F⋅x−W2⋅d=0\begin{aligned} \sum \tau &= 0 \\ F \cdot x - W_2 \cdot d &= 0 \end{aligned}
3. We obtain a system with unknowns F and x:
{F=W1+W2W2⋅d=F⋅x⇒{F=m1g+m2gm2g⋅d=(m1g+m2g)⋅x\begin{cases} F = W_1 + W_2 \\ W_2 \cdot d = F \cdot x \end{cases} \Rightarrow \begin{cases} F = m_1 g + m_2 g \\ m_2 g \cdot d = (m_1 g + m_2 g) \cdot x \end{cases}
Applying the data we get:
F=686 Nx=0.85 m\begin{aligned} \htmlClass{text-primary}{F} &\htmlClass{text-primary}{= 686 \text{ N}} \\ \htmlClass{text-primary}{x} &\htmlClass{text-primary}{= 0.85 \text{ m}} \end{aligned}
When a body undergoes uniform circular motion, there is a net force that ties it to the center of the trajectory. This is the centripetal force. It can be the tension of a string, the friction of a road... etc.
Fc=m⋅ac⇒Fc=m⋅v2R⇒Fc=mω2RF_c = m \cdot a_c \quad \Rightarrow \quad F_c = m \cdot \frac{v^2}{R} \quad \Rightarrow \quad F_c = m \omega^2 R
Knowing that a ball of mass m=8 kg, attached to a string, undergoes uniform circular motion in a vertical circle, with ω=3 rad/s\omega = 3 \text{ rad/s}. Calculate the tension of the string at points A and B in the figure
Vertical string tension statement
Approach at point A
1. Free-body diagram on the ball:
The acceleration that the mass experiences is the normal (centripetal) acceleration, towards the center of the trajectory.
Point A free body diagram
2. Newton's 2nd Law equation: Y-axis only.
Recall that ac=v2R=ω2Ra_c = \frac{v^2}{R} = \omega^2 R
W+T=m⋅acT=m⋅ω2R−m⋅g\begin{aligned} W + T &= m \cdot a_c \\ T &= m \cdot \omega^2 R - m \cdot g \end{aligned}
3. Applying the data from the problem and solving:
T=m(ω2R−g)=65.6 NT = m(\omega^2 R - g) = \htmlClass{text-primary}{65.6 \text{ N}}
Approach at point B
1. Free-body diagram of the ball:
In this case, T and W are in opposite directions.
Point B free body diagram
2. Newton's 2nd Law equation:
T−W=m⋅acT=m⋅ω2R+m⋅g\begin{aligned} T - W &= m \cdot a_c \\ T &= m \cdot \omega^2 R + m \cdot g \end{aligned}
3. Applying the data and solving:
T=m(ω2R+g)=222.4 NT = m(\omega^2 R + g) = \htmlClass{text-primary}{222.4 \text{ N}}