What Rocket Thrust Accelerates This Sled?

Prior to manned space flights, rocket sleds were used to test aircraft, missile equipment, and physiological effects on humans at high accelerations. Rocket sleds consisted of a platform mounted on one or two rails and propelled by several rockets. Calculate the magnitude of force exerted by each rocket, called its thrust, T, for the four-rocket propulsion system shown below. The sled’s initial acceleration is $49 {m/s}^{2},$ the mass of the system is 2,100 kg, and the force of friction opposing the motion is 650 N.

An accelerating right-facing rocket sled is shown on tracks. Four thrust force vectors are behind the sled and are pointing to the right. An acceleration vector is above the sled and is pointing to the right. A friction force vector is in front of the sled and is pointing to the left. A Newton vector is below the sled and is pointing upward. A weight vector is also below the sled and is pointing downward. A free body diagram illustrates the force vectors.

Figure 4.7

Work it out on paper first. Then open the solution one step at a time, and stop as soon as you can finish on your own.

Strategy
The system of interest is the rocket sled. Although forces act vertically on the system, they must cancel because the system does not accelerate vertically. This leaves us with only horizontal forces to consider. We’ll assign the direction to the right as the positive direction. See the free-body diagram in Figure 4.8.
Solution
We start with Newton’s second law and look for ways to find the thrust T of the engines. Because all forces and acceleration are along a line, we need only consider the magnitudes of these quantities in the calculations. We begin with

$F_{net} = m a,$
4.11

where $F_{net}$ is the net external force in the horizontal direction. We can see from Figure 4.8 that the engine thrusts are in the same direction (which we call the positive direction), whereas friction opposes the thrust. In equation form, the net external force is

$F_{net} = 4 T - f .$
4.12

Newton’s second law tells us that F_net= ma, so we get

$m a = 4 T - f .$
4.13

After a little algebra, we solve for the total thrust 4T:

$4 T = m a + f,$
4.14

which means that the individual thrust is

$T = \frac{m a + f}{4} .$
4.15

Inserting the known values yields

$T = \frac{(2100 {kg)(49 m/s}^{2}) + 650 N}{4} = 2.6 \times 10^{4} N .$
4.16

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Strategy

Solution