1.)A 16.0 kg loudspeaker is suspended 2.00 mbelow the ceiling by two 3.90 m long

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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