Question 5.7: A ramp, a pulley, and two boxes. A box of mass mA = 10.0 kg ......

(a) For both cases (i) and (ii), Newton’s second law for the y direction (perpendicular to the plane) is the same:

F_{\mathrm{N}}-m_{\mathrm{A}} g \cos \theta=m_{\mathrm{A}} a_y=0

since there is no y motion. So

F_{\mathrm{N}}=m_{\mathrm{A}} g \cos \theta.

Now for the x motion. We consider case (i) first for which \Sigma F= ma gives

m_{\mathrm{A}} g \sin \theta-F_{\mathrm{T}}-F_{\mathrm{fr}}=m_{\mathrm{A}} a_x

We want a_x = 0 and we solve for F_{\mathrm{T}} since F_{\mathrm{T}} is related to m_{\mathrm{B}} (whose value we are seeking) by F_{\mathrm{T}} = m_{\mathrm{B}}  g (see Fig. 5-9c).Thus

m_{\mathrm{A}} g \sin \theta-F_{\mathrm{fr}}=F_{\mathrm{T}}=m_{\mathrm{B}}  g.

We solve this for m_{\mathrm{B}} and set F_{\mathrm{fr}} at its maximum value \mu_{\mathrm{s}} F_{\mathrm{N}}=\mu_{\mathrm{s}} m_{\mathrm{A}} g \cos \theta to find the minimum value that m_{\mathrm{B}} can have to prevent motion (a_x = 0):

\begin{aligned}m_{\mathrm{B}} & =m_{\mathrm{A}} \sin \theta-\mu_{\mathrm{s}} m_{\mathrm{A}} \cos \theta \\& =(10.0 \mathrm{~kg})\left(\sin 37^{\circ}-0.40 \cos 37^{\circ}\right)=2.8 \mathrm{~kg} .\end{aligned}

Thus if m_{\mathrm{B}} < 2.8 kg, then box A will slide down the incline.

Now for case (ii) in Fig. 5-9b, box A being pulled up the incline. Newton’s second law is

m_{\mathrm{A}} g \sin \theta+F_{\mathrm{fr}}-F_{\mathrm{T}}=m_{\mathrm{A}} a_x=0.

Then the maximum value m_{\mathrm{B}} can have without causing acceleration is given by

F_{\mathrm{T}}=m_{\mathrm{B}} g=m_{\mathrm{A}} g \sin \theta+\mu_{\mathrm{S}} m_{\mathrm{A}} g \cos \theta

or

\begin{aligned}m_{\mathrm{B}} & =m_{\mathrm{A}} \sin \theta+\mu_{\mathrm{s}} m_{\mathrm{A}} \cos \theta \\& =(10.0 \mathrm{~kg})\left(\sin 37^{\circ}+0.40 \cos 37^{\circ}\right)=9.2 \mathrm{~kg} .\end{aligned}

Thus, to prevent motion, we have the condition

2.8 \mathrm{~kg}<m_{\mathrm{B}}<9.2 \mathrm{~kg}.

(b) If m_{\mathrm{B}} = 10.0 kg and \mu_k = 0.30, then m_{\mathrm{B}} will fall and m_{\mathrm{A}} will rise up the plane (case ii). To find their acceleration a, we use \Sigma F = ma for box A:

m_{\mathrm{A}} a=F_{\mathrm{T}}-m_{\mathrm{A}} g \sin \theta-\mu_{\mathrm{k}} F_{\mathrm{N}}.

Since m_{\mathrm{B}} accelerates downward, Newton’s second law for box B (Fig. 5-9c) tells us m_{\mathrm{B}}  a=m_{\mathrm{B}}  g  – F_{\mathrm{T}}, \text { or } F_{\mathrm{T}}=m_{\mathrm{B}}  g-m_{\mathrm{B}}  a, and we substitute this into the equation above:

m_{\mathrm{A}}  a=m_{\mathrm{B}}  g-m_{\mathrm{B}} a-m_{\mathrm{A}} g \sin \theta-\mu_{\mathrm{k}} F_{\mathrm{N}} .

We solve for the acceleration a and substitute F_{\mathrm{N}}=m_{\mathrm{A}} g \cos \theta , and then m_{\mathrm{A}}=m_{\mathrm{B}} = 10.0 kg, to find

\begin{aligned}a & =\frac{m_{\mathrm{B}} g-m_{\mathrm{A}} g \sin \theta-\mu_{\mathrm{k}} m_{\mathrm{A}} g \cos \theta}{m_{\mathrm{A}}+m_{\mathrm{B}}} \\& =\frac{(10.0 \mathrm{~kg})\left(9.80 \mathrm{~m} / \mathrm{s}^2\right)\left(1-\sin 37^{\circ}-0.30 \cos 37^{\circ}\right)}{20.0 \mathrm{~kg}} \\& =0.079 g=0.78 \mathrm{~m} / \mathrm{s}^2 .\end{aligned}

NOTE It is worth comparing this equation for acceleration a with that obtained in Example 5-5: if here we let θ = 0, the plane is horizontal as in Example 5-5, and we obtain a=\left(m_{\mathrm{B}} g-\mu_{\mathrm{k}} m_{\mathrm{A}} g\right) /\left(m_{\mathrm{A}}+m_{\mathrm{B}}\right) just as in Example 5-5 .