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What rocket thrust accelerates this sled?

Prior to manned space flights, rocket sleds were used to test aircraft, missile equipment, and physiological effects on human subjects at high speeds. They consisted of a platform that was mounted on one or two rails and propelled by several rockets. Calculate the magnitude of force exerted by each rocket, called its thrust T size 12{T} {} , for the four-rocket propulsion system shown in [link] . The sled’s initial acceleration is 49 m/s 2 , size 12{"49"" m/s" rSup { size 8{2} } } {} the mass of the system is 2100 kg, and the force of friction opposing the motion is known to be 650 N.

A sled is shown with four rockets, each producing the same thrust, represented by equal length arrows labeled as vector T pushing the sled toward the right. Friction force is represented by an arrow labeled as vector f pointing toward the left on the sled. The weight of the sled is represented by an arrow labeled as vector W, shown pointing downward, and the normal force is represented by an arrow labeled as vector N having the same length as W acting upward on the sled. A free-body diagram is also shown for the situation. Four arrows of equal length representing vector T point toward the right, a vector f represented by a smaller arrow points left, vector N is an arrow pointing upward, and the weight W is an arrow of equal length pointing downward.
A sled experiences a rocket thrust that accelerates it to the right. Each rocket creates an identical thrust T size 12{T} {} . As in other situations where there is only horizontal acceleration, the vertical forces cancel. The ground exerts an upward force N size 12{N} {} on the system that is equal in magnitude and opposite in direction to its weight, w size 12{w} {} . The system here is the sled, its rockets, and rider, so none of the forces between these objects are considered. The arrow representing friction ( f size 12{f} {} ) is drawn larger than scale.


Although there are forces acting vertically and horizontally, we assume the vertical forces cancel since there is no vertical acceleration. This leaves us with only horizontal forces and a simpler one-dimensional problem. Directions are indicated with plus or minus signs, with right taken as the positive direction. See the free-body diagram in the figure.


Since acceleration, mass, and the force of friction are given, we start with Newton’s second law and look for ways to find the thrust of the engines. Since we have defined the direction of the force and acceleration as acting “to the right,” we need to consider only the magnitudes of these quantities in the calculations. Hence we begin with

F net = ma size 12{F rSub { size 8{"net"} } = ital "ma"} {} ,

where F net size 12{F rSub { size 8{"net"} } } {} is the net force along the horizontal direction. We can see from [link] that the engine thrusts add, while friction opposes the thrust. In equation form, the net external force is

F net = 4 T f size 12{-F rSub { size 8{"net"} } =4T-f} {} .

Substituting this into Newton’s second law gives

F net = ma = 4 T f size 12{F rSub { size 8{"net"} } = ital "ma"=4T-f} {} .

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

4 T = ma + f size 12{4T= ital "ma"+f} {} .

Substituting known values yields

4 T = ma + f = ( 2100 kg ) ( 49 m/s 2 ) + 650 N size 12{4T= ital "ma"+f= \( "2100"" kg" \) \( "49 m/s" rSup { size 8{2} } \) +"650"" N"} {} .

So the total thrust is

4 T = 1.0 × 10 5 N size 12{4T=1 "." "04" times "10" rSup { size 8{5} } " N"} {} ,

and the individual thrusts are

T = 1.0 × 10 5 N 4 = 2.6 × 10 4 N size 12{T= { {1 "." "04" times "10" rSup { size 8{5} } " N"} over {4} } =2 "." 5 times "10" rSup { size 8{4} } " N"} {} .


The numbers are quite large, so the result might surprise you. Experiments such as this were performed in the early 1960s to test the limits of human endurance and the setup designed to protect human subjects in jet fighter emergency ejections. Speeds of 1000 km/h were obtained, with accelerations of 45 g size 12{g} {} 's. (Recall that g size 12{g} {} , the acceleration due to gravity, is 9 . 80 m/s 2 size 12{9 "." "80 m/s" rSup { size 8{2} } } {} . When we say that an acceleration is 45 g size 12{g} {} 's, it is 45 × 9 . 80 m/s 2 size 12{"45"´9 "." "80 m/s" rSup { size 8{2} } } {} , which is approximately 440 m/s 2 size 12{"440 m/s" rSup { size 8{2} } } {} .) While living subjects are not used any more, land speeds of 10,000 km/h have been obtained with rocket sleds. In this example, as in the preceding one, the system of interest is obvious. We will see in later examples that choosing the system of interest is crucial—and the choice is not always obvious.

Questions & Answers

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Practice Key Terms 7

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