Forces F1 and F2 act on a mass m = 0.67 kg, as shown in the diagram below. The m

Question

Forces F1 and F2 act on a mass m = 0.67 kg, as shown in the diagram below. The magnitudes of the forces and angles are as follows: F1 = 1.5 N, 1 = 22 degrees F2 = 4.1 N, 2 = 58 degrees

What are the x and y components of the force F1?
(  ,  ) N

What are the x and y components of the force F2?
(  ,  ) N

What are the x and y components of the sum of these two forces (the resultant force)?
(  ,  ) N

What is the magnitude of the resultant force?
N

What are the x and y components of the resulting acceleration of the mass?
(  ,  m/s2

What is the magnitude of the acceleration?
m/s2

Tries 0/2

2) The force F = (60, -80) N acts on a mass of 2 kg. At time t = 0 s, the mass is initially at rest.

In which quadrant (1-4) is the force?

What is the angle (in degrees) of the force, measured counter-clockwise from the +x-axis?

What is the magnitude of the force (in N)?

What are the (x,y) components of the acceleration vector (in m/s2)?
(  ,  )

What are the (x,y) components of the velocity vector (in m/s) after 1.7 seconds?
(  ,  )

Explanation / Answer

F1x = F1*cos22 = 1.39N

F1y = F1*sin22 = 0.56 N

F2x = F2*cos58 = 4.1*cos58 = 2.17 N

F2y = F2*sin58 = 4.1*sin58 = 3.48 N

Rx = F1x +F2x = 3.56 N

Ry = F1y +F2y = 4.04 N

R = sqrt(Rx^2 +Ry^2) = 5.38 N

ax = Rx/m = 3.56/0.67 = 5.31 m/s^2

ay = Ry/m = 4.04/.67 = 6.02 m/s^2

a = sqrt(ax^2 +ay^2) = 8.03 m/s^2
---------------------------------

a) 4th

b) tan^-1(-80/60) = 306.87

c) F = sqrt(80^2+60^2) = 100 N

d) ax = Fx/m = 60/2 = 30 m/s^2

ay = -80/2 = -40 m/s^2

e) Vx = ux + ax*t = 0 + (30*1.7) = 51 m/s

Vy = uy + ayt = 0 + (-40*1.7) = -68 m/s

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