Engineering Science 2

[匿名]

2009-10-15 12:13 am

5. A woman of mass 70.0 kg weighs herself in an elevator.
a. If she wants to weigh less, should she weigh herself when accelerating upward or
downward?
b. When the elevator is not accelerating, what does the scale read (i.e. what is the
normal force that the scale exerts on the woman)?
c. When the elevator is accelerating upward at
2.00 m/s2, what does the scale read?

6. A heavy box (mass 25 kg) is dragged along the floor by a kid at a 30° angle to
the horizontal with a force of 80 N (which is the maximum force the kid can
apply).
a. Draw the free body diagram.
b. What is the normal force FN?
c. Does the normal force decrease or increase as the angle of pull increases? Explain.
d. Assuming no friction, what is the acceleration of the box?
e. Assuming it begins at rest, what is its speed after ten seconds?
f. Is it possible for the kid to lift the box by pulling straight up on the rope?
g. In the absence of friction, what is the net force in the x-direction if the kid pulls at a
30° angle?
h. In the absence of friction, what is the net force in the x-direction if the kid pulls at a
45° angle?
i. In the absence of friction, what is the net force in the xdirection
if the kid pulls at a 60° angle?
j. The kid pulls the box at constant velocity at an angle of 30°.
What is the coefficient of kinetic friction μK between the box
and the floor?
k. The kid pulls the box at an angle of 30°, producing an
acceleration of 2 m/s2. What is the coefficient of kinetic friction μK between the box and
the floor?

回答 (1)

Prof. Physics

2009-10-16 5:50 am

✔ 最佳答案

5.a. She should weigh herself when accelerating downwards.

b. The scale reads her own weight, i.e. 700 N

c. By Newton's 2nd law of motion,

R - mg = ma

R = mg + ma = (70)(10 + 2) = 840 N

6.b. Fn = Mg - Rsin30* = (25)(10) - (80)(1/2) = 210 N

c. It decreases when the angle of pull increases, as the angle of pull increases, the vertical component of the applied force acting upward is larger, so the numerical value of normal force can become smaller in order to obtain equilibrium in vertical direction.

d. By F = ma

80cos30* = 25a

Acceleration, a = 2.77 ms^-2

e. By v = u + at

v = 0 + (2.77)(10)

Speed after ten seconds, v = 27.7 m/s

f. It is not possible for the kid to lift the box, as the maximum applied force is smaller than the weight of the box.

g. Net force in x-direction = Rcos30* = 69.3 N

h. Net force in x-direction = Rcos45* = 56.6 N

i. Net force in x-direction = Rcos60* = 40 N

j. Friction = 69.3 N

By umg = 69.3

u(250) = 69.3

Coefficient of kinetic friction, u = 0.277

k. By Newton's 2nd law of motion,

Rcos30* - f = ma

80cos30* - f = (25)(2)

Friction, f = 19.3 N

By umg = f

u(250) = 19.3

Coefficient of kinetic friction, u = 0.077

參考: Physics king

👍 0 👎 0