Question #aecf3

2 Answers
Jan 27, 2018

Answer:

#20N# to #45^@# with the x-axis

Explanation:

Well, they are in opposite directions, and the #45^@# one has more force, so it wins.

Thus, the resultant force is #30N-10N=20N#.

Since the #45^@# is stronger, the resultant force with be #20N# with a direction of #45^@# with the x-axis.

Jan 27, 2018

Answer:

#39.74 N "at "41.3^@# to the x axis.

Explanation:

I will assume that the #30^@ "and 45^@# angles are both counterclockwise from the +x axis. (Or both could be clockwise -- my result would not be different if so.)

The x axis component of the 10 N force would be
#F_"1x" = 10 N*cos30^@ = 8.66 N#
The y axis component of the 10 N force would be
#F_"1y" = 10 N*sin30^@ = 5.0 N#

The x axis component of the 30 N force would be
#F_"2x" = 30 N*cos45^@ = 21.21 N#
The y axis component of the 30 N force would be
#F_"2y" = 30 N*sin45^@ = 21.21 N#

The x component of the resultant is
#8.66 N +21.21 N = 29.87 N#
The y component of the resultant is
#5.0 N +21.21 N = 26.21 N#

The magnitude of the resultant is
#sqrt ((29.87 N)^2 + (26.21 N)^2) = 39.74 N#

The direction of the resultant is #"tan"^-1(26.21/29.87) = 41.3^@#

I hope this helps,
Steve