Question 16.SP.2: A sled is jet-propelled along a straight track by a force P ...
A sled is jet-propelled along a straight track by a force P that increases linearly with time according to P = kt, where k is a constant. The coefficient of sliding friction between the sled runners and the track is μ_k, the coefficient of static friction is μ_s, and the mass of the sled is m. Determine (a) the time at which the tip of the rocket begins to rotate downward, (b) the acceleration of the sled at this instant. Neglect loss of mass due to fuel consumption and assume that the sled will slide before it tips.
STRATEGY: Since you are given a force, use Newton’s second law to find the acceleration required for the rocket to begin rotating forward. You can then find the time using P = kt.
MODELING: Choose the sled as your system and model it as a rigid body. The rocket force must overcome the static friction force before it begins moving. Define this time to be t_0. Figure 1 shows a free-body diagram when the motion is impending. In this case, both of the friction forces are set equal to the maximum allowable friction force μ_sN. Free-body and kinetic diagrams for when the sled is about to tip are shown in Fig. 2. Just as the sled starts to tip, the normal force on the rear of the sled goes to zero.
ANALYSIS: Using Fig. 1 and applying Newton’s second law in the y-and x-directions gives
+\uparrow \Sigma F_y=m \bar{a}_y: \quad N_A + N_B-m g=0 \quad \text { or } \quad N_A + N_B=m g
\stackrel{+}{\rightarrow} \Sigma F_x=m \bar{a}_x: \quad k t_0 – \left(\mu_s N_A + \mu_s N_B\right)=0
or
k t_0=\mu_s\left(N_A + N_B\right)=\mu_s m g (1)
Now that you know when the sled begins to slide, you can determine the time it will start to tip using Fig. 2.
From this diagram, you can apply Newton’s second law in the x and y directions and sum moments about any point. If you choose to take moments about G, you find
\stackrel{+}{\rightarrow} \Sigma F_x=m \bar{a}_x: \quad k\left(t – t_0\right) – \mu_k N=m \bar{a} (2)
+\uparrow \Sigma F_y=m \bar{a}_y: \quad N-m g=0 (3)
+\circlearrowleft \Sigma M_G=\bar{I} \alpha: \quad N d-\mu_k N b – k\left(t – t_0\right) c=0 (4)
Solving Eqs. (1), (2), (3), and (4) for t_0, t, N, \bar{a}, you find N = mg, t_0=\mu_s m g / k and
\begin{gathered}t=\frac{m g\left(d + c \mu_s – b \mu_k\right)}{k c} \\\bar{a}=\frac{g\left(d – c \mu_k-b \mu_k\right)}{c}\end{gathered}
REFLECT and THINK: Rather than taking moments about G, you could have chosen any other point. For example, for moments about A, you have
+\circlearrowleft \Sigma M_A=\bar{I} \alpha + m \bar{a} d: \quad m g d – k\left(t – t_0\right)(b + c)=-m \bar{a} b
Using this equation rather than Eq. (4) will give you the same answer. To check the assumption that the sled slides before it tips, you would need to use Fig. 1 and show that both N_A and N_B are positive for the given value of P=k t_0.