A uniform beam is mechanically attached to a wall supposed by a wire and at angl

Question

A uniform beam is mechanically attached to a wall supposed by a wire and at angle of 35 degrees to the horizontal. The mass of the beam is 20kg and an alien of mass 10kg is sitting a third of the way along the beam. Write down Newton's 2nd law equation for torque and use it to find the tension in the wire. Be sure to justify your reasoning. Write down the Newton's second law equations for force and calculate the x and y components of the force exerted by the wall. What is the magnitude of the force exerted by the wall and what angle does it make to the horizontal?

Explanation / Answer

Here,

mass of beam , m = 20 Kg

theta = 35 degree

m1 = 10 Kg

1)

let the tension in the string is T

balancing the torque about the balance point on wall

T * sin(theta) * L - m * g * L/2 - m1 * 2L/3 *g = 0

T * sin(35) - 20 * 9.8/2 - 10 * 2/3 * 9.8 = 0

T = 285 N

the tension in the wire is 285 N

2)

for the horizontal and vertical force

in horizontal direction

Fx - T * cos(theta) = 0

Fx - 285 * cos(35) = 0

Fx = 233.3 N

the x component of the force is 233.3 N

for the vertical direction

Fy + T * sin(theta) = m1 * g + m * g

Fy + 285 * sin(35) = (10 + 30) * 9.8

Fy = 130.6 N

the vertical component force is 130.6 N

3)

for the net force on the wall

angle of force exerted by the wall = arctan(Fy/Fx)

angle of force exerted by the wall = arctan(130.6/233.3)

angle of force exerted by the wall = 29.2 degree

for

magnitude of force exerted by the wall= sqrt(Fx^2 + Fy^2)

magnitude of force exerted by the wall = sqrt(130.6^2 + 233.3^2)

magnitude of force exerted by the wall = 267.4 N

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