m₁ 1112 -12- A uniform rod of mass m 10.0 cm and 2 "1 = = 163 g and length L = 100.0 cm is attached to the wall with a pin as shown. Cords are attached to the rod at the 90.0 cm mark, passed over pulleys, and masses of m. = 281 g and m₂ m1 157 g are attached. Your TA asks you to determine the following. = (a) The position r3 on the rod where you would suspend a mass m3 = 200 g in order to balance the rod and keep it horizontal if released from a horizontal position. In addition, for this case, what force (magnitude and direction) does the pin exert on the rod? Use standard angle notation to determine the direction of the force the pin exerts on the rod. Express the direction of the force the pin exerts on the rod as the angle 0 F, measured with respect to the positive x-axis (counterclockwise is positive and clockwise is negative). = 3 0.4395 Fp 0F = 0.735 -90 m N ° (b) Let's now remove the mass m3 and determine the new mass m 4 you would need to suspend from the rod at the position r = 20.0 cm in order to balance the rod and keep it horizontal if released from a horizontal position. In addition, for this case, what force (magnitude and direction) does the pin exert on the rod? Express the direction of the force the pin exerts on the rod as the angle OF, measured with respect to the positive x-axis (counterclockwise is positive and clockwise is negative). m4 = 0.4395 F р F = 1.61 = 90 kg N ° (c) Let's now remove the mass m 4 and determine the mass m 5 you would suspend from the rod in order to have a situation such that the pin does not exert a force on the rod and the location г from which you would suspend this mass in order to balance the rod and keep it horizontal if released from a horizontal position. m5 = 0.275 r5 × kg = See if you can write a condition for rotational equilibrium statement that will allow you to determine the position on the rod where we need to suspend the mass m5 in order for the rod to remain stationary when released from a horizontal position. m

College Physics
11th Edition
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Chapter1: Units, Trigonometry. And Vectors
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m₁
1112
-12-
Transcribed Image Text:m₁ 1112 -12-
A uniform rod of mass m
10.0 cm and 2
"1
=
= 163 g and length L = 100.0 cm is attached to the wall with a pin as shown. Cords are attached to the rod at the
90.0 cm mark, passed over pulleys, and masses of m. = 281 g and m₂
m1
157 g are attached. Your TA asks you to determine the following.
=
(a) The position r3 on the rod where you would suspend a mass m3 = 200 g in order to balance the rod and keep it horizontal if released from a horizontal
position. In addition, for this case, what force (magnitude and direction) does the pin exert on the rod? Use standard angle notation to determine the
direction of the force the pin exerts on the rod. Express the direction of the force the pin exerts on the rod as the angle 0 F, measured with respect to the
positive x-axis (counterclockwise is positive and clockwise is negative).
=
3 0.4395
Fp
0F
=
0.735
-90
m
N
°
(b) Let's now remove the mass m3 and determine the new mass m 4 you would need to suspend from the rod at the position r = 20.0 cm in order to balance
the rod and keep it horizontal if released from a horizontal position. In addition, for this case, what force (magnitude and direction) does the pin exert on the
rod? Express the direction of the force the pin exerts on the rod as the angle OF, measured with respect to the positive x-axis (counterclockwise is positive
and clockwise is negative).
m4 = 0.4395
F
р
F
=
1.61
=
90
kg
N
°
(c) Let's now remove the mass m 4 and determine the mass m 5 you would suspend from the rod in order to have a situation such that the pin does not exert
a force on the rod and the location г from which you would suspend this mass in order to balance the rod and keep it horizontal if released from a horizontal
position.
m5
=
0.275
r5
×
kg
= See if you can write a condition for rotational equilibrium statement that will allow you to determine the position on the rod where we need to suspend
the mass m5 in order for the rod to remain stationary when released from a horizontal position. m
Transcribed Image Text:A uniform rod of mass m 10.0 cm and 2 "1 = = 163 g and length L = 100.0 cm is attached to the wall with a pin as shown. Cords are attached to the rod at the 90.0 cm mark, passed over pulleys, and masses of m. = 281 g and m₂ m1 157 g are attached. Your TA asks you to determine the following. = (a) The position r3 on the rod where you would suspend a mass m3 = 200 g in order to balance the rod and keep it horizontal if released from a horizontal position. In addition, for this case, what force (magnitude and direction) does the pin exert on the rod? Use standard angle notation to determine the direction of the force the pin exerts on the rod. Express the direction of the force the pin exerts on the rod as the angle 0 F, measured with respect to the positive x-axis (counterclockwise is positive and clockwise is negative). = 3 0.4395 Fp 0F = 0.735 -90 m N ° (b) Let's now remove the mass m3 and determine the new mass m 4 you would need to suspend from the rod at the position r = 20.0 cm in order to balance the rod and keep it horizontal if released from a horizontal position. In addition, for this case, what force (magnitude and direction) does the pin exert on the rod? Express the direction of the force the pin exerts on the rod as the angle OF, measured with respect to the positive x-axis (counterclockwise is positive and clockwise is negative). m4 = 0.4395 F р F = 1.61 = 90 kg N ° (c) Let's now remove the mass m 4 and determine the mass m 5 you would suspend from the rod in order to have a situation such that the pin does not exert a force on the rod and the location г from which you would suspend this mass in order to balance the rod and keep it horizontal if released from a horizontal position. m5 = 0.275 r5 × kg = See if you can write a condition for rotational equilibrium statement that will allow you to determine the position on the rod where we need to suspend the mass m5 in order for the rod to remain stationary when released from a horizontal position. m
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