In a hockey game two defencemen simultaneously hit an opposing forward, who has a mass of 80 kg. One defenceman exerts a force on the forward of 475 N, 25° W of N. The other defenceman exerts a force on the forward of 40 N, 60° E of N. Determine the net force acting on the forward and the forward’s acceleration.

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aristocles aristocles · Answer 1 · 2018-11-14T02:44:31+01:00

here two forces are acting simultaneously

[tex]F_1 = 475 N[/tex] 25^0 W of N

[tex]F_2 = 40 N[/tex] 60^0 E of N

now we will write the two forces in components form

[tex]F_1 = -475 sin25 \hat i + 475 cos25 \hat j[/tex]

[tex]F_1 = -200.7 \hat i + 430.5 \hat j[/tex]

[tex]F_2 = 40sin60 \hat i + 40 cos60 \hat j[/tex]

[tex]F_2 = 34.6 \hat i + 20 \hat j[/tex]

now net force on it will be

[tex]F = F_1 + F_2 [/tex]

[tex]F = (-200.7 + 34.6) \hat i + (430.5 + 20)\hat j[/tex]

[tex]F = -166.1 \hat i + 450.5 \hat j[/tex]

so net force is of magnitude

[tex]F = \sqrt{166.1^2 + 450.5^2}[/tex]

[tex]F = 480.1 N[/tex]

direction is given as

[tex]\theta = tan^{-1}\frac{166.1}{450.5} = 20.2 degree[/tex]

so net force is 480.1 N at 20.2 degree W of N

now the acceleration is given by Newton's II law

[tex]F = ma[/tex]

[tex]480.1 = 80 a[/tex]

[tex]a = 6 m/s^2[/tex]