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Question (17.38, Hibbeler Dynamics SI Edition):

A force of $P = 300\,\mathrm N$ is applied to the $60\:\text{kg}$ cart. Determine the reactions at both the wheels at A and both the wheels at B. Also, what is the acceleration of the cart? The mass center of the cart is at G.

In the question above, we can clearly see that the force is pulling towards the left, and hence the acceleration of $4.33\: \text{m/s}^2$ occurs towards the left. But while solving for the reaction forces at the wheels of the cart, why do we consider the force $ma$ acting towards the right? This question is from an exercise that has a lot of questions involving pseudo/d'Alembert forces.

I'm attaching the link to the correct solution to the problem. https://study.com/academy/answer/a-force-of-p-300n-is-applied-to-the-60-kg-cart-determine-the-reactions-at-both-the-wheels-at-a-and-b-also-what-is-the-acceleration-of-the-cart-the-mass-center-of-the-cart-is-at-g.html

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  • $\begingroup$ By including the pseudo-force you are now solving a statics problem. $\endgroup$ – Farcher Dec 21 '19 at 6:03
  • $\begingroup$ @Farcher I understand that, but will the effects of this pseudo force even be felt in real life? What I mean is, say when we are going up in an elevator, we feel 'heavier'. Is the pseudo force used in the question actually going to have an effect like this? $\endgroup$ – Bhavya M. Dec 21 '19 at 9:49
  • $\begingroup$ The pseudo force is very well felt in real life. A great example is when you are in a car taking a turn. You experience a centrifugal force outward while the car is under a centripetal force inwards. $\endgroup$ – Sam Dec 21 '19 at 10:41
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Questions such as this can be solved with or without the use of pseudo force; it all depends on the frame of reference with which you are solving the question.

The wheel is connected to the cart by the axle which is accelerating towards the left and it taking the wheels along with it. From the frame of reference of the wheels(non-inertial frame), the wheels are experiencing a pseudo force in the opposite direction of its motion ie towards the right.

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  • $\begingroup$ Thank you, that makes much more sense. $\endgroup$ – Bhavya M. Dec 21 '19 at 10:42
  • $\begingroup$ You may accept this answer and close the questions if your doubt has been clarified. :) $\endgroup$ – Sam Dec 21 '19 at 10:45

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