Two children tie two strings at the same point of a trolley and pull with a force of 20 Newtons in a horizontal plane, so between the strings is an angle of 90o

Two children tie two strings at the same point of a supermarket trolley and pull with a force of 20 Newtons in a horizontal plane, so between the strings is an angle of 90^o.

a) Represent the two forces on a scale of 1 cm = 10 Newtons;

b) Compare the two force vectors;

c) If its speed is constant?

Solution

a) We represent the support of the forces through two lines that intersect in a point and form an angle of 90^o between them (Fig.1). Starting from intersection point 0, each right-handed segment with a length of 2 cm is represented on each support. We note the two vectors F1 and F2.

b) The vectors F1 and F2 have equal numerical values ​​F1 = F2 = 20 Newtons, they have the same application point, but have different directions and meanings. Therefore, the vectors F1 and F2 are not equal F1 ≠ F2.

c) The trolley speed is constant if the effect of the F1 and F2 forces is offset by the effect of the Ff friction force between the wheels and the asphalt. The force that would produce the same effect as the F1 and F2 forces is their resultant (R):

R = F1 + F2

It can be found by using the parallelogram rule for vectors: a parallelogram is constructed which has the vectors F1 and F2 as sides (Fig.2), leading through the F1 tip a parallel to F2 and through the F2 tip a parallel to F1. The parallelogram diagonal that starts at point 0 is the resultant of F1 and F2 forces. We can assume that two forces are exerted on the trolley: the force R (replacing F1 and F2) and the friction force Ff. Since the trolley's speed remains constant, it means that the effects of the two forces R and Ff are compensated, so their result is null:

R + Ff = 0

This is only possible if the forces R and Ff have the same direction, opposite senses and equal numerical values ​​so that R - Ff = 0 (Fig.2). It is measured in Fig. 1 the length of the diagonal and it is 2.8 cm. Taking into account the chosen scale (1 cm = 10 Newtons), it results that R = 28 Newtons. Therefore, the friction force is:

 Ff = 28 Newtons

Mechanics

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