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jeyben [28]
3 years ago
6

A force of 38 newtons is acting on an object making an angle of 65º with the horizontal. What is the value component of the forc

e acting along the y-axis?
(A)4.98 newtons
(B)16.8 newtons
(C)34.4 newtons
(D)50.8 newtons
(E)96.1 newtons
Physics
2 answers:
RoseWind [281]3 years ago
6 0
The component of the force along the y-axis is given by
F_y = F \sin \theta
where
F is the magnitude of the force
\theta is the angle between the direction of the force and the horizontal (x-axis)

In our problem, F=38 N and \theta=65^{\circ}, therefore the y-component of the force is
F_y = F \sin \theta=(38 N)(\sin 65^{\circ})=34.4 N
Therefore, the correct answer is C).
Pachacha [2.7K]3 years ago
5 0
The answer is C.................

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A bucket of mass m is hanging from the free end of a rope whose other end is wrapped around a drum (radius R, mass M) that can r
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Explanation:

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Applying newton's second law on the downward movement of the bucket

mg - T = ma

On the drum , a torque of TR will be acting which will create an angular acceleration of α in it . If I be the moment of inertia of the drum

TR = Iα

TR = Ia/ R

T =  Ia/ R²

Replacing this value of T in the other equation

mg - T = ma

mg - Ia/ R²  = ma

mg =  Ia/ R² +ma

a ( I/ R² +m)= mg

a = mg / ( I/ R² +m)

mg - T = ma

mg - ma  = T

mg - m x mg / ( I/ R² +m) = T

mg - m²g / ( I/ R² +m ) = T

mg - mg / ( 1 + I / m R² ) = T

b ) T =  Ia/ R²

I = TR² / a

c ) Moment of inertia of hollow cylinder

I = 1/2  M ( R² - R² / 4 )

= 3/4 x 1/2 MR²

= 3/8 MR²

I / R² = 3/8 M

a = mg / ( I/ R² +m)

a = mg / ( 3/8 M + m )

T =  Ia/ R²

= 3/8 MR² x mg / ( 3/8 M + m ) x 1 /R²

= \frac{3mMg}{(3M +8m)}

7 0
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