1) A horizontal rope is tied to a 60 kg box on frictionless ice. a)What is the t

Question

1) A horizontal rope is tied to a 60 kg box on frictionless ice.

a)What is the tension in the rope if the box is at rest? N

b)What is the tension in the rope if the box moves at a steady 4.0 m/s ? N

c)What is the tension in the rope if the box has vx = 4.0 m/s and ax = 5.0 m/s2 N

2)A loudspeaker of mass 20.0 kg is suspended a distance of h = 2.20 m below the ceiling by two cables that make equal angles with the ceiling. Each cable has a length of l = 3.30 m

a)What is the tension T in each of the cables?

Use 9.80 m/s2 for the magnitude of the acceleration due to gravity

3) A chandelier with mass m is attached to the ceiling of a large concert hall by two cables. Because the ceiling is covered with intricate architectural decorations (not indicated in the figure, which uses a humbler depiction), the workers who hung the chandelier couldn't attach the cables to the ceiling directly above the chandelier. Instead, they attached the cables to the ceiling near the walls. Cable 1 has tension T1 and makes an angle of 1 with the ceiling. Cable 2 has tension T2 and makes an angle of 2 with the ceiling.

a)

Find an expression for T1 , the tension in cable 1, that does not depend on T2 .

Express your answer in terms of some or all of the variables m , 1 , and 2 , as well as the magnitude of the acceleration due to gravity g . You must use parentheses around 1 and 2 , when they are used as arguments to any trigonometric functions in your answer. T1= ????

4) A gymnast of mass 70.0 kg hangs from a vertical rope attached to the ceiling. You can ignore the weight of the rope and assume that the rope does not stretch. Use the value 9.81m/s2 for the acceleration of gravity.

a)Calculate the tension T in the rope if the gymnast hangs motionless on the rope. N=?

b)Calculate the tension T in the rope if the gymnast climbs the rope at a constant rate. N=?

c)Calculate the tension T in the rope if the gymnast climbs up the rope with an upward acceleration of magnitude 0.500 m/s2 . N= ?

d) Calculate the tension T in the rope if the gymnast slides down the rope with a downward acceleration of magnitude 0.500 m/s2 . N=?

Explanation / Answer

1) a steady velocity means acceleration is zero.
f = ma when a=0, f=0 too
there tension is zero.
now there is an acceleration
apply f=ma

= 60*5 =
f=300N

2)

The force due to gravity is 9.8 * 20= 196 N.

This mean each cable experiences 98 N of vertical tension (235.2 / 2).

For the x-direction, we know that the length of each cable is 3.3 m and the height is 2.2m. This means we can solve for the angle (sin() = 2.2/3.3). So = °. The vertical component of the tension is Ty = Tsin(). Since we know that Ty = 196 N, we can solve for T. T is just 98 / sin),

you should find the angel you can get the answer as i gave u formula

3)

Fnet = 0
T1sintheta1 + T2sintheta2 = mg

In the horizontal direction, the forces are also balanced:
Fnet = 0
T1costheta1 = T2costheta2
T2 = T1(costheta1)/(costheta2)

Using this result in the first equation:
T1sintheta1 + T2sintheta2 = mg
T1sintheta1 + T1(costheta1)(sintheta2)/(costheta2) = mg
T1sintheta1 + T1(costheta1)(tantheta2) = mg
T1(sintheta1 + (costheta2)(tantheta2)) = mg

T1 = mg / (sintheta1 + (costheta2)(tantheta2))

T1 = mg*cos theta2 / sin(theta1+theta2)

4)a) accel =0, and the forces are T and W where W = 70kg or W=686.7 N

T-W=0 =>T=W =686.7 N

b) climbing at constant speed still means a=0, same answer as above

c)Fnet = ma
T - mg = ma
T = mg + ma
T = 70(9.81) + 70(0.500)
T = 721.7 N

d)
Fnet = ma
T - mg = ma
T = mg + ma
T = 70(9.81) + 70(-0.500)
T = 651.7 N

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