please show your work so that I can see how you did it. This is one of the pract

Question

please show your work so that I can see how you did it. This is one of the practice questions for the lab I am about to do using a force table, if that helps. Thank You!

Prelab Exercise: Below, is a diagram illustrating the forces that two people exert on heavy box. (a) Add the two force vectors together, to find the total force exerted by the two people. What is the magnitude and direction of the total force. (b) Suppose that a third force were exerted on the box in such a way that the total force on the box is zero. What is the magnitude and direction of this third force? 120 N 80 N 75%

Explanation / Answer

One method of doing this is to break both vectors into their x components and y components

for 120 N

Fx = 120cos60 = 120*0.5 = 60N

Fy = 120sin60 = 120*sqrt(3)/2 = 103.92 N

for 80 N

Fx = -80cos75 = -20.64N (negative sign means -x direction)

Fy = 80sin75 = 80*0.965 = 77.2N

net resultant x component Fx = 60-20.64 = 39.36 N

net resultant y component Fy = 103.92 + 77.2 = 181.12 N

F = sqrt(Fx^2 + Fy^2) = sqrt(39.36^2 + 181.12^2) = 185.35 N

let direction angle be theta

tan(theta) = Fy/Fx = 181.12/39.36 = 4.602

theta = 77.74o from x-axis

part b)

Magnitude of third force will be same with direction 180o opposite to resultant of above 2 forces i.e. 180+77.74 = 257.74o from x axis

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