Solution

The force diagram is analysed just as before. The only difference is in the appearance of an additional force $T$. Along the parallel direction, however this is merely one more force to include in the net force: $$ \begin{eqnarray} mg\sin\theta + T &=& m a \\ \Rightarrow a &=& g\sin\theta + \frac{T}{m}\\ &=& 9.8\sin30^\circ + \frac{15}{2} \\ &=& 4.9 + 7.5 = 12.4m/s^2 \end{eqnarray} $$

In the perpendicular direction, the tension has no effect. Again, using the fact that $a_\perp = 0$, we can conclude $F_n = mg\cos30^\circ = 16.97N$.

Solution

Begin by drawing 2 separate free-body diagrams, one for each object: For simplicity we only included the horizontal forces because the motion is clearly purely horizontal. The vertical forces like the normal forces and the weight cancel out.

Now treat the diagrams as if they are two separate problems and apply the 2nd Law to each: $$ \begin{eqnarray} T&=&m_1 a \\ F-T &=& m_2 a \end{eqnarray} $$ From these 2 equations, we can solve for the 2 unknowns \(a\) and \(T\). The fastest way to do it is by adding the equations together (usually the action and reaction forces, which are \(T\) and \(-T\) here will cancel out): $$ \begin{eqnarray} F &=& m_1 a + m_2 a = (m_1+m_2)a \\ \Rightarrow a &=& \frac{F}{m_1+m_2} \\ &=& \frac{60N}{1kg+2kg} = 20m/s^2 \end{eqnarray} $$

Substituting \(a=20m/s^2\) back into either one of the equations earlier gives the tension: $$ \begin{eqnarray} T&=&m_1 a = (1kg)(20m/s^2) = 20N \end{eqnarray} $$

Solution

We will begin by decomposing the force \(F\): Noting that both \(a\) and \(T\) are horizontal, we draw the free-body diagrams of each object, including only the horizontal forces (we will look at the verticl forces in the follow-up question here):

One can see that the force diagrams are essentially the same as the previous example, the only difference is that \(F\) is now replaced by \(F\cos\theta\). The calculation of \(a\) and \(T\) therefore proceed exactly as before. The 2nd Law on each of the two force diagrams give: $$ \begin{eqnarray} T&=&m_1 a \\ F\cos \theta-T &=& m_2 a \end{eqnarray} $$ Adding the equations give: $$ \begin{eqnarray} F\cos\theta &=& m_1 a + m_2 a = (m_1+m_2)a \\ \Rightarrow a &=& \frac{F\cos\theta}{m_1+m_2} \\ &=& \frac{30N \cos 20^\circ}{1kg+2kg} = 9.40 m/s^2 \end{eqnarray} $$

Substituting \(a=20m/s^2\) back into either one of the equations earlier gives the tension: $$ \begin{eqnarray} T&=&m_1 a = (1kg)(9.40m/s^2) = 9.40N \end{eqnarray} $$

Solution

Just like before, we begin by drawing the free-body diagrams, one for each object, but this time focus on the vertical forces (because the normal forces are vertical): Note that the force \(F\) acting on \(m_2\) has been decomposed as in the last problem, and only the vertical component \(F\sin\theta\) is kept. Because there is no motion vertically, we know the vertical acceleration is zero, i.e. \(a_y = 0\). This implies all the vertical forces must cancel out.

For \(m_1\): $$ \begin{eqnarray} F_{n1} &=& m_1 g = (1kg)(9.8m/s^2) = 9.8N \end{eqnarray} $$

For \(m_2\): $$ \begin{eqnarray} F_{n2} + F\sin\theta &=& m_2 g \\ \Rightarrow F_{n2} &=& m_2 g - F\sin\theta \\ &=& (2)(9.8) - 30\sin 20^\circ \\ &=& 19.6N - 10.26N = 9.34N \end{eqnarray} $$

Solution

For any problems involving friction, it is always a good idea to figure out the normal force first (because it is need to find friction). This problem was actually tackled in Chapter 04a, but here is a reminder.

First we decompose the vector $F_1$, then we collect all the vertical forces in a force diagram (ignoring the horizontal forces for now). Since there is no vertical acceleration, the vertical forces must all cancel out and we get: $$ \begin{eqnarray} F_n &=& mg + F_1 \sin 50^\circ = (2)(9.8) + 10 \sin 50^\circ \\ &=& 19.60 + 7.66N = 27.26N \end{eqnarray} $$

From this we could calculate the kinetic friction: $$ \begin{eqnarray} f &=& \mu F_n \\ &=& (0.12)(27.26N) = 3.27N \end{eqnarray} $$ The horizontal forces can now be used to find the acceleration: $$ \begin{eqnarray} 10\cos 50^\circ - f &=& ma \\ a &=& \frac{10\cos 50^\circ - f}{m} \\ &=& \frac{6.43 - 3.27}{2} = 1.58m/s^2 \end{eqnarray} $$

Solution

Solution