1. A worker opens a 1.10 m wide door by pushing on it with a force of 36.5 N dir
ID: 1791742 • Letter: 1
Question
1. A worker opens a 1.10 m wide door by pushing on it with a force of 36.5 N directed perpendicular to its surface. HINT (a) What magnitude torque (in N · m) does she apply about an axis through the hinges if the force is applied at the center of the door? N · m (b) What magnitude torque (in N · m) does she apply at the edge farthest from the hinges? N · m
2.Find the net torque on the wheel in the figure below about the axle through O perpendicular to the page, taking a = 5.00 cm and b = 19.0 cm. (Indicate the direction with the sign of your answer. Assume that the positive direction is counterclockwise.)
N · m
3.A uniform 33.0-kg beam of length
= 4.95 m
is supported by a vertical rope located
d = 1.20 m
from its left end as in the figure below. The right end of the beam is supported by a vertical column.
(a) Find the tension in the rope.
N
(b) Find the force that the column exerts on the right end of the beam. (Enter the magnitude.)
N
4.Each of the following objects has a radius of 0.161 m and a mass of 2.07 kg, and each rotates about an axis through its center (as in this table) with an angular speed of 36.1 rad/s. Find the magnitude of the angular momentum of each object.
(a) a hoop
kg · m2/s
(b) a solid cylinder
kg · m2/s
(c) a solid sphere
kg · m2/s
(d) a hollow spherical shell
kg · m2/s
5.
A student holds a spinning bicycle wheel while sitting motionless on a stool that is free to rotate about a vertical axis through its center (see the figure below). The wheel spins with an angular speed of 16.1 rad/s and its initial angular momentum is directed up. The wheel's moment of inertia is 0.190 kg · m2 and the moment of inertia for the student plus stool is 2.70 kg · m2.
HINT
(a)
Find the student's final angular speed (in rad/s) after he turns the wheel over so that it spins at the same speed but with its angular momentum directed down.
rad/s
(b)
Will the student's final angular momentum be directed up or down?
updown
Explanation / Answer
Given
width of the door w = 1.10 m
F = 36.5 N
we know that the torque T = R*F sin theta
here theta = 90 degrees
a) when r = 1.1/2
T = r*F
= (1.1/2)*36.5 N m
T = 20.075 N m
b) r= 1.1 m
T = 1.1*36.5 N m
T = 40.15 N m
2. T1 = r*F sin theta
T1 = b*F sin theta
T1 = - 0.19*10 sin90 = 1.9 N.m
T2 = r*F sin theta
T2 = b*F sin theta
T2 = - 0.19*9 sin90 = 1.71 N.m
T3 = r*F sin theta
T3 = a*F sin theta
T3 = 0.05*12 sin30 = 0.3 N.m
T = T3 -(T1+T2)
T = 0.3 -(1.9+1.71) N.m
T = -3.31 N.m
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