Let** F1, F2, F3, F4**, and ** F5** be the forces exerted by the five persons A, B, C, D and E respectively on the cart to pull the cart. These are all vector quantities.

Let the unit vector (force in Newtons) in the direction of East be denoted by **i**. Let the unit vector in North direction be denoted by ** j**. Then:

when all of them pull the cart:

vector Sum* F1 + F2 + F3 + F4 + F5 = 100 * 3* * i = 300 **i **Newtons. --- (1)

when A does not pull the cart,

vector sum *F2 + F3 + F4 + F5 =* 100 * 1 * (** - i **) = - 100 **i ** Newtons. --- (2)

Subtract the two above equations from one another: **F1** = 400 **i **.

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It is not clear when person B stops pulling, Whether A is pulling the cart or not.. Let us assume that A , C, D and E are pulling the cart with a = 24 m/s/s in North.

vector sum *F1 + F3 + F4 + F5 =* 100 * 24 * *j* = 2400 *j* Newtons --- (3)

eq (1) - (3) gives : **F2** = 300 **i -** 2,400 **j **Newtons.

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If only A and B pull the cart, then the net force on the cart is :

vector sum *F1* + **F2** = 700 *i *- 2, 400 **j **Newtons

The acceleration of the cart will then be : ** a** =* F* / mass = 7** i** - 24** j** m/sec²

* magnitude of acceleration = √(7² + 24²) = 25 m/sec²*