This question has several parts that must be completed sequentially. If you skip a part of the question, you will not re
Posted: Fri May 06, 2022 11:18 am
This question has several parts that must be completed
sequentially. If you skip a part of the question, you will not
receive any points for the skipped part, and you will not be able
to come back to the skipped part. Tutorial Exercise A 470-N uniform
rectangular sign 4.00 m wide and 3.00 m high is suspended from a
horizontal, 6.00-m-long, uniform, 145-N rod as indicated in the
figure below. The left end of the rod is supported by a hinge and
the right end is supported by a thin cable making a 30.0° angle
with the vertical. (Assume the cable is connected to the very end
of the 6.00-m-long rod, and that there are 2.00 m separating the
wall from the sign.) A rectangular sign hangs along the bottom of a
rod connected to a hinge on a wall on its left. A cable with
tension T at the right end of the rod connects to the wall above
the hinge and makes a 30.0° angle with the vertical. The right end
of the sign is at the right end of the rod. (a) Find the (magnitude
of the) tension T in the cable. (b) Find the horizontal and
vertical components of the force exerted on the left end of the rod
by the hinge. (Take up and to the right to be the positive
directions.) Step 1 The free-body diagram of the sign–rod
combination is shown below. A rectangular sign hangs down from a
6.00 m horizontal rod. The right side of the sign is at the right
end of the rod and the center of the sign is 4.00 m to the right of
the left end of the rod. Six forces act on the sign-rod
combination. A force of magnitude Rx acts horizontally rightwards
on the left end of the rod. A force of magnitude Ry acts vertically
upwards on the left end of the rod. A force vector Wr acts
vertically downwards on the rod 3.00 m from the left end of the
rod. A force vector Ws acts vertically downwards on the center of
the sign horizontally 4.00 m from the left end of the rod. A force
of magnitude Tx = T sin(30.0°) acts horizontally leftwards on the
right end of the rod. A force of magnitude Ty = T cos(30.0°) acts
vertically upwards on the right end of the rod. The tension in the
cable and the reaction force exerted on the left end of the rod by
the hinge have been resolved into horizontal and vertical
components. The weight of the rod w with arrowr acts at the center
of the rod, (6.00 m)/2 = 3.00 m from the wall, and the weight of
the sign w with arrows acts at the center of the sign, a total of
6.00 m − (4.00 m)/2 = 4.00 m from the wall. Note that the height of
the sign does not figure into any of the calculations because it is
not the lever arm of any torque and because we are given the weight
of the sign. Consider a rotation axis perpendicular to the page,
passing through the left end of the rod. We apply the second
condition of equilibrium, Σ𝜏 = 0, to this system. We have + T cos
30.0° 6.00 m − N 3.00 m − N m = 0 or T = ✕ 103 N · m 6.00 m cos
30.0° = N.
sequentially. If you skip a part of the question, you will not
receive any points for the skipped part, and you will not be able
to come back to the skipped part. Tutorial Exercise A 470-N uniform
rectangular sign 4.00 m wide and 3.00 m high is suspended from a
horizontal, 6.00-m-long, uniform, 145-N rod as indicated in the
figure below. The left end of the rod is supported by a hinge and
the right end is supported by a thin cable making a 30.0° angle
with the vertical. (Assume the cable is connected to the very end
of the 6.00-m-long rod, and that there are 2.00 m separating the
wall from the sign.) A rectangular sign hangs along the bottom of a
rod connected to a hinge on a wall on its left. A cable with
tension T at the right end of the rod connects to the wall above
the hinge and makes a 30.0° angle with the vertical. The right end
of the sign is at the right end of the rod. (a) Find the (magnitude
of the) tension T in the cable. (b) Find the horizontal and
vertical components of the force exerted on the left end of the rod
by the hinge. (Take up and to the right to be the positive
directions.) Step 1 The free-body diagram of the sign–rod
combination is shown below. A rectangular sign hangs down from a
6.00 m horizontal rod. The right side of the sign is at the right
end of the rod and the center of the sign is 4.00 m to the right of
the left end of the rod. Six forces act on the sign-rod
combination. A force of magnitude Rx acts horizontally rightwards
on the left end of the rod. A force of magnitude Ry acts vertically
upwards on the left end of the rod. A force vector Wr acts
vertically downwards on the rod 3.00 m from the left end of the
rod. A force vector Ws acts vertically downwards on the center of
the sign horizontally 4.00 m from the left end of the rod. A force
of magnitude Tx = T sin(30.0°) acts horizontally leftwards on the
right end of the rod. A force of magnitude Ty = T cos(30.0°) acts
vertically upwards on the right end of the rod. The tension in the
cable and the reaction force exerted on the left end of the rod by
the hinge have been resolved into horizontal and vertical
components. The weight of the rod w with arrowr acts at the center
of the rod, (6.00 m)/2 = 3.00 m from the wall, and the weight of
the sign w with arrows acts at the center of the sign, a total of
6.00 m − (4.00 m)/2 = 4.00 m from the wall. Note that the height of
the sign does not figure into any of the calculations because it is
not the lever arm of any torque and because we are given the weight
of the sign. Consider a rotation axis perpendicular to the page,
passing through the left end of the rod. We apply the second
condition of equilibrium, Σ𝜏 = 0, to this system. We have + T cos
30.0° 6.00 m − N 3.00 m − N m = 0 or T = ✕ 103 N · m 6.00 m cos
30.0° = N.