i need help, am sick. The system shown in the following figure is in static equi
ID: 2133272 • Letter: I
Question
i need help, am sick.
The system shown in the following figure is in static equilibrium and the angle 0 = 32°. Given that the mass m1 is 8.40 kg and the coefficient of static friction between mass m1 and the surface on which it rests is 0.310, what is the minimum mass that m2 can have for which the system will still remain in equilibrium? kg Two known forces F1 = 3.6 N at 77degree below the +x axis and F2 = 3.6 N at 38degree above the -x axis are acting on an object, (a) What is the magnitude of the third force F3 so that the acceleartion of the object is zero? N In serving, a tennis player accelerates a 57-gram tennis ball horizontally from rest to a speed of 30 m/s. Assuming the acceleration is uniform when the racquet applies a force over a distance of 2.10 m, what is the magnitude of the force exerted on the ball? N Jane and John, with masses of 47 kg and 63 kg, respectively, stand on a frictionless surface 14 m apart. John pulls on a connecting rope, giving Jane an acceleration of 1.5 m/s2 toward him. If the pulling force is applied constantly, how far Jane has travelled when they meet? mExplanation / Answer
7) min mass m2 so it will be going down the slope
m1 g sin theta - m2g - friction = (m1 + m2)*0
friction = u m1 g cos theta
8.4*sin(32) - m2 - 0.31*8.4*cos(32)=0
m2 = -0.31*8.4*cos(32) + 8.4*sin(32)=2.24 kg
8)
F3x = -F1x - F2x = -3.6*cos(77) - 3.6*cos(38)=-3.65 N
F3y = -3.6*sin(77) -3.6*sin(38)=-5.72
so F3 = sqrt(3.65^2 + 5.72^2)=6.79 N
9)
v^2 = v0^2 + 2 a x
a =v^2/(2x)
F = ma= 57.0E-3*30^2/(2*2.21)=11.6 N
10) that means forces of rope = 47*1.5
John will also feel the same force ans will feel an a = 47*1.5/63
so xjane = 1/2*1.5*t^2
x john = 14 - 0.5*47*1.5/63*t^2
xjohn = xjane
1/2*1.5*t^2= 14 - 0.5*47*1.5/63*t^2
t=3.27
so xjane = 0.5*1.5*3.27^2=8.02 m
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