the tensions T1 and T2 in the strings must satisfy which of the following relations A Tl T2 B T1 gt T2 C T1 lt T2 D Tl T2 mg Answer As T2 is more vertical it is supporting more of the weight of the ball The ID: 545194
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Slide1Slide2
1
. A ball of mass m is suspended from two strings of unequal length as shown above. The magnitudes of
the tensions
T1 and T2 in the strings must satisfy which of the following relations?(A) Tl = T2 (B) T1 > T2 (C) T1 < T2 (D) Tl + T2 = mgSlide3
Answer
As T2 is more vertical, it is supporting more of the weight of the ball. The
horizontal components
of T1 and T2 are equal.CSlide4Slide5
Answer
2. Normal force is perpendicular to the incline, friction acts up, parallel to the incline (opposite
the motion
of the block), gravity acts straight down.ESlide6
Which
of the following correctly indicates the magnitudes of the forces acting up and down the incline?
(A) 20 N down the plane, 16 N up the plane
(B) 4 N down the plane, 4 N up the plane(C) 0 N down the plane, 4 N up the plane
(D) 10 N down the plane, 6 N up the planeSlide7
Answer
The component of the weight down the plane is 20 N sin
30. The
net force is 4 N, so the friction force up the plane must be 4 N less than 10 N.DSlide8
A
plane 5 meters in length is inclined at an angle of 37°, as shown above. A block of weight 20 N is placed
at the
top of the plane and allowed to slide down.
The
magnitude of the normal force exerted on the block by the plane is
(A) greater than 20 N
(B) greater than zero but less than 20 N
(C) equal to 20 N
(D) zeroSlide9
Answer
The
weight component perpendicular to the plane is 20 N sin 37o. To get
equilibrium perpendicular to the plane, the normal force must equal this weight component, which must be less than 20 N.BSlide10Slide11
a.
b. f =
μ
N where N = m1g cos θ gives μ
= f /(m
1
g cos
θ
)
c. constant velocity means
Σ
F = 0 where
Σ
F
external
= m
1
g sin
θ
+ m
2
g sin
θ
– f – 2f – Mg = 0
solving for M gives M = (m
1 + m2) sin θ – (3f)/gd. Applying Newton’s second law to block 1 gives Σ F = m1g sin θ – f = m1a which gives a = g sin θ – f/m1