The two boxcars A and B have a weight of 24 000 lb and 33 000 lb, respectively.
ID: 1426841 • Letter: T
Question
The two boxcars A and B have a weight of 24 000 lb and 33 000 lb, respectively. They are freely coasting down the incline when the brakes are applied to all the wheels of car A. The coefficient of kinetic friction between the wheels of A and the tracks is mu_k = 0.6. The wheels of car B are free to roll. Neglect their mass in the calculation. Suggestion: Solve the problem by representing single resultant normal forces acting on A and B, respectively. (Figure 1) Determine the magnitude of the force in the coupling C between the two cars. Express your answer to three significant figures and include the appropriate units.Explanation / Answer
Let me solve this in MKS units since I am comfortable with it. Nevertheless, you can finally convert the answer into what ever units you want by multiplying with the appropriate factor.
Ma = 24000 lb = 10886.217 Kg
Mb = 33000 lb = 14968.548 Kg
inclination angle = 5 degrees
On block A Force in the direction of motion (due to gravity ) is Ma xg xSin = 10886.217x 9.81x Sin 5 = 9307.692 N
Friction = Normal reaction force x k = Ma x g x Cos x k = 10886.217 x 9.81 x Cos 5 x 0.6 = 63832.444N
Let the compressive force in the coupling C be T (In appeared to my intuition that the force in the coupling C will be compressive, if at all we get a negative value for T, it just means that the force is tensile. So, no worries.)
Since, we assumed the force in C to be compressive, it will push the block A downward,
So, net acceleration of block A downwards = net forces on A / mass of A
a = ( Ma xg xSin - Ma x g x Cos x k + T )/ Ma
a = ( 9307.692 - 63832.444 + T )/ 10886.217
a = ( T - 54524.752 ) / 10886.217
Now, coming to block B, Force in the direction of motion (due to gravity ) is Mb xg xSin = 14968.548x 9.81x Sin 5
= 12798.076 N
No, friction on B since the wheels of B are smooth.
Since, we assumed the force in coupling C to be compressive, it will push the block B upwards
Therefore net acceleration of block B = net forces on B / mass of B
a = ( Mb xg xSin - T ) / 14968.548
a = ( 12798.076 - T ) / 14968.548
The accelerations of the two blocks need to be the same, since the vehicle is a connected body going together.
Hence equating the accelerations we get,
a = ( T - 54524.752 ) / 10886.217 = ( 12798.076 - T ) / 14968.548
( T - 54524.752 ) / 10886.217 = ( 12798.076 - T ) / 14968.548
Solving for T in this eqation gives,
14968.548 T - 816156367.5 = 139322632.5 - 10886.217 T
25854.765 T = 955479000
T = 36955.625
Note the positive value which means that our assumption of compreesive force in the coupling is true.
Further note that this is in SI units, which means T = 36955.625 Newton
or T= 36955.625 x 0.224809 = 8307.955 lbf
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