An object with a mass of #4 kg# is acted on by two forces. The first is #F_1= < 6 N , -2 N># and the second is #F_2 = < -3 N, 3 N>#. What is the object's rate and direction of acceleration?

1 Answer
Apr 9, 2016

Answer:

The acceleration, #vec(a) = 1/4[(3), (1)]m/s^2#
and the magnitude is:
#|vec(a)| = 1/4sqrt[(3)^2+(1)^2]N=1/4sqrt(10)m/s^2#
You can write the acceleration, #vec(a)= sqrt10/4[(3/sqrt(10)), (1/sqrt(10))]m/s^2#
Where# [(3/sqrt(10)), (1/sqrt(10))]=# is the unit vector of #vec(a)#

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Explanation:

Given: Forces on an object of mass, #m=4kg#
#vec(F_1)=[(6), (-2)]N# and #vec(F_2)=[(-3), (3)]N#

Required: Object acceleration, #vec(a)#

Solution Strategy:
A) Draw a "Free Body Diagram" and add the two force vector
B) Use Newton's law #sum_l(F_xhati + F_yhatj)=m(a_xhati+a_yjhat)#
That is:
1) #" "sum_l F_x = ma_x" "# #and" "# 2) #" "sum_l F_y = ma_y#

A) See Image Below

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The FBD yield:
#F_(x)=F_(1_x)+F_(2_x)=6-3= 3N#
#F_(y)=F_(1_y)+F_(2_y)=-2+3= 1N #
Thus the net force, #vec(F_"net") = [(3), (1)]N#

B) Newton Law:
From equation 1) we have:
1) #F_(x)=ma_x; a_x=3/4 m/s^2#
#F_(y)=ma_y; a_y=1/4 m/s^2#

Thus the net acceleration, #vec(a) = 1/4[(3), (1)]m/s^2#
and the magnitude is:
#|vec(a)| = 1/4sqrt[(3)^2+(1)^2]N = 1/4sqrt(10)m/s^2#