991 Words4 Pages

Jean W Charles 10/22/12
1. How hard must you pull on the third rope to keep the knot from moving?
Put the knot at 0,0 and the 3 unit force along the x axis. The 5 unit force goes into the second quadrant
3 units * cos(ᶱ) = + 3 units
5 units * cos(120o) = -2.5 units.
Resultant = +3 - 2.5 = 0.5 units.
Sum the forces vertically
3*sin(ᶱ) = 0
5*sin(120o) = 4.33 units
Resultant + 4.33 vertically
Find where you should be.
F = sqrt(Fx^2 + Fy^2)
F = sqrt(4.33^2 + 0.5^2)
F = sqrt(19)
F = 4.35 N
So you must pull with 4.35 units of force.
Now for the angle the angle = 180 + Tan-1(vertical/horizontal) = 180 + tan-1(4.33/0.5) = 263.4 o from the + x axis.*…show more content…*

4. A highway curve of radius 500 m is designed for traffic moving at a speed of 90 km/hr. What is the correct banking angle of the road? In order that the driving speed is as safe as possible, we need not depend on any traction force. Thus the only forces involved are the road's normal force N and the automobile's weight m*g. The car is traveling in horizontal motion, and thus all vertical forces must add up to zero. N*cos(theta) = m*g And we can solve for N: N = m*g/cos(theta) Apply Newton's 2nd law for the radial direction: N*sin(theta) = m*a Substitute: m*g*sin(theta)/cos(theta) = m*a cancel m, simplify trigonometry combination: g*tan(theta) = a The acceleration is centripetal acceleration, and is thus equal to: a = v^2/r Substitute: g*tan(theta) = v^2/r Solve for theta: theta = arctan(v^2/(r*g)) Data: v:=90*10/36 m/s; g:=9.8 N/kg; r:=500 m; Result: theta = 7.269 degrees 5. How much work does an elevator motor do to lift a 1000 kg elevator a height of 100m? How much power must the motor supply to do this in 50 s? Use the following formula to find Fm (necessary to find ): Fnet = ma (m = mass of elevator, and a = acceleration) Since the elevator's moving at constant speed, acceleration =0. So, Fnet = 0 Fnet = Fm - Fg = 0 Solve for Fm: Fm = Fg = mg Fm = (1000 kg)(9.8 m/s2) = 9800 N Now you can solve for work: W = F d = (9800 N) (100 m) = 980,000J (Nm = J) Now to find power:

4. A highway curve of radius 500 m is designed for traffic moving at a speed of 90 km/hr. What is the correct banking angle of the road? In order that the driving speed is as safe as possible, we need not depend on any traction force. Thus the only forces involved are the road's normal force N and the automobile's weight m*g. The car is traveling in horizontal motion, and thus all vertical forces must add up to zero. N*cos(theta) = m*g And we can solve for N: N = m*g/cos(theta) Apply Newton's 2nd law for the radial direction: N*sin(theta) = m*a Substitute: m*g*sin(theta)/cos(theta) = m*a cancel m, simplify trigonometry combination: g*tan(theta) = a The acceleration is centripetal acceleration, and is thus equal to: a = v^2/r Substitute: g*tan(theta) = v^2/r Solve for theta: theta = arctan(v^2/(r*g)) Data: v:=90*10/36 m/s; g:=9.8 N/kg; r:=500 m; Result: theta = 7.269 degrees 5. How much work does an elevator motor do to lift a 1000 kg elevator a height of 100m? How much power must the motor supply to do this in 50 s? Use the following formula to find Fm (necessary to find ): Fnet = ma (m = mass of elevator, and a = acceleration) Since the elevator's moving at constant speed, acceleration =0. So, Fnet = 0 Fnet = Fm - Fg = 0 Solve for Fm: Fm = Fg = mg Fm = (1000 kg)(9.8 m/s2) = 9800 N Now you can solve for work: W = F d = (9800 N) (100 m) = 980,000J (Nm = J) Now to find power:

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