3-8 Part C: Breaking Up a Force Into Its Components (In this part we will work b

Question

3-8 Part C: Breaking Up a Force Into Its Components (In this part we will work backward. In particular we will use the hypothesis that forces add like vectors to break a single force into components along perpendicular directions.) C-1 Remove all inasses from the table and set up a single pulley at the 240 mark with a 400 g mass suspended from it. c-2 Set up pulleys at the 0 and 90° marks as illustrated in Figure 3-6 and deter- mine the masses required to balance the single given force. Keep the angles fixed but adjust the masses and determine the experimental range for each one. 90° Soo 400 g 240° FIGURE 3-6. Given force and pulley positions for Part C. C-3 The two forces you determined in step C-2 must have a resultant equal to the equilibrant for the single given force. Therefore, they must be equivalent to the equilibrant; i.e., their vector sum is the equilibrant. Furthermore, since they are perpendicular, they must actually be the conlponents of the equilibrant-400 g @60° Use this fact to determine the magnitudes of the two forces analytically C-4 Compare your results from steps C-2 and C.3. (Display them prominently in a table to make the comparison easy) Do your analytical values for the components of the equilibrant fall in the ranges that you determined experimentally?

Explanation / Answer

Solution C-2

given force= mg=0.4g=3.92 N

Component of the given force in perpendicular direction

in downwards direction= 0.4gcos(30)=0.3464g N

in left= 0.4*g*sin(30)=0.2*g N

let unknow masses be m1 @ 0 degrees and m2 @ 90 degrees

upwards force= m2g

balancing with downwards force we get: m2g=0.3464 g

m2=0.3464 kg

force at right= m1g

balancing with force in left: m1g=0.2 g

m1=0.2 kg

Solution C-3

let the two forces in upwards and right direction be F1 and F2

then these must be the vertical and horizontal components of given force, 3.92 N

3.92cos(30)=F1 and 3.92sin(30)=F2

F1=3.3948 N; F2=1.96 N

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