A 37.4 kg beam is attached to a wall with a hinge and its far end is supported by a cable. The angle between the beam and the cable is 90o. If the beam is inclined at an angle of θ = 16.5° with respect to horizontal.
a) What is the horizontal component of the force exerted by the hinge on the beam? (Use the `to the right' as + for the horizontal direction.)
b)What is the magnitude of the force that the beam exerts on the hinge?
a) What is the horizontal component of the force exerted by the hinge on the beam? (Use the `to the right' as + for the horizontal direction.)
b)What is the magnitude of the force that the beam exerts on the hinge?
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Resolving mg into two components mg cos θ and mg sin θ,
and since tension T is perpendicular to the beam and since the mg acts at the center of the beam ,
mg cos θ = 2 T and the force on the hinge perpendicular to the beam = T
The force along the beam = mg sin θ
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Answer for a)
mg sin θ* cos θ = 37.4 *9.8 sin 16.5 *cos 16.5=99.8 N
Answer for b)
mg sin θ = 37.4 *9.8 sin 16.5=104.1 N
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and since tension T is perpendicular to the beam and since the mg acts at the center of the beam ,
mg cos θ = 2 T and the force on the hinge perpendicular to the beam = T
The force along the beam = mg sin θ
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Answer for a)
mg sin θ* cos θ = 37.4 *9.8 sin 16.5 *cos 16.5=99.8 N
Answer for b)
mg sin θ = 37.4 *9.8 sin 16.5=104.1 N
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