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Question

..Three forces of magnitudes F1=4.0N, F2=6.0N, and F3=8.0N are applied to a block of mass m=2.0kg, initially at rest, at angles shown on the diagram. (Figure 1)In this problem, you will determine the resultant (net) force by combining the three individual force vectors. All angles should be measured counterclockwise from the positive x axis (i.e., all angles are positive).

A) Calculate the magnitude of the resultant force Fx r= -->F1+-->F2+-->F3 acting on the block.
Express the magnitude of the resultant force in newtons to two significant figures.

B) What angle does -->Fr make with the positive x axis?
Express your answer in degrees to two significant figures.

oubaas · Answer

Fx = 4*cos 25 + 6*cos 35 -8 = 0.54
Fy = 4*sin 25-6*sin35° = -1.75 
F = √Fx^2+Fy^2 = √0.54^2+1.75^2 = 1.8 N

angle = 360-arctan -Fy/Fx = 287.1°

oldschool · Answer

A) Calculate the magnitude of the resultant force Fx r= -->F1+-->F2+-->F3 acting on the block. 
Sum horizontal components:
4*cos25 + 6*cos325 + 8*cos180 = 0.54

Sum vertical components:
4*sin25 + 6*sin325 + 8*sin180 = -1.75
F² = 0.54²+1.75² = 1.8 A) 
at arctan(-1.75/0.54) = -73° = 290°   B)

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