1.)A 16.0 kg loudspeaker is suspended 2.00 mbelow the ceiling by two 3.90 m long
ID: 1334492 • Letter: 1
Question
1.)A 16.0 kg loudspeaker is suspended 2.00 mbelow the ceiling by two 3.90 m long cables that angle outward at equal angles.What is the tension in each of the cables?
The figure shows the force acting on a 3.0 kg object as it moves along the x-axis. The object is at rest at the origin at t=0s.
a. What is its acceleration at
6s ?
b.What is its velocity at
6s ?
3.A 3000
kg truck is parked on a 18 slope.How big is the friction force on the truck? The coefficient of static friction between the tires and the road is 0.90.
please help !!
Explanation / Answer
here,
the horizontal angle between the loudspeaker and ceiling is , theta = arcsin(2/3.9)
theta = 30.85 degree
and,
mass of the loudspeaker , m = 16 kg
the force of gravity (downwards) is mg, ie 16 * 9.8 = 156.8 N
Now let the tension in both the strings be T, therefore, the total upward force will be 2Tsin(30.85)
Therefore , 156.8 = 2* T* sin(30.85)
T = 154.84 N
the tension in each cable is 154.84 N
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