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Question

A 134 N uniform beam with a length of 2.2 m is attached to a pivot at one end, a cable at the other, and another cable at a point a distance of one-fourth the length of the beam away from the point where the other cable attaches. The cable on the left in the image has a magnitude of tension of 180 N, and the angles in the image are θ1= 29∘, θ2= 35∘, and φ= 68∘

a) Determine the magnitude of tension in the cable on the right, in N.

b) Determine the horizontal component of the force at the pivot, in N. Use a coordinate system where a force component to the right is positive.

c) Determine the vertical component of the force at the pivot, in N. Use a coordinate system where a force component in the upwards direction is positive.

Diagram: https:// usb06ca.theexpertta.com/images/bkbfmbtc.tpy.png

Answer 1

I'm using a picture for this problem with the bar φ = 68° above the horizon and the two cables are θ₁= 29°, θ₂= 35° above the horizon. There are four forces on the bar. A force from the pivot, a force from each cable, and gravity. The force from the pivot is the hardest to determine so it is a good idea to measure torques from the pivot.

a)

Problem outline

1) graph torques

2) calculate torques

3) plug into the sum of torques

4) solve with algebra

Remember to use the equation

τ = r F sinθ

we need to graph r and F in the standard position (tail at origin) on the same axis. θ will go from r to F with counter-clockwise being positive. There are lots of "quick" ways to calculate torques your class may have used before and problems like this will catch you trying to take shortcuts.

Let the left cable corresponding to θ₁ be the first torque. The force vector should be in the third quadrant. It should be 29° below the negative x-axis.

r = L = 2.2m

F = T₁ = 180N

θ = (90°-68°) + 90 + 29° = 141°

τ₁ = (2.2m)(180N)sin(141°)

This is a positive number and should cause a the bar to rotate counter-clockwise in the absence of the other torques. That means the signs are right.

Let the right cable corresponding to θ₂ be the second torque. The force vector should be in the fourth quadrant 35° below the positive x-axis.

r = L3/4 = 1.65 m

F = T₂ = F₂

θ = -35° - 68° = -103°

τ₂ = (0.55m)(F₂)sin(-103°)

This is a negative number and should cause a the bar to rotate clockwise in the absence of the other torques. That means the signs are again right.

The last torque comes from gravity. Gravity acts at the center of mass.

r = L/2 = 1.1 m

F = F_g = 134N

θ = -90°-68° = -158°

τ_g = (1.1m)(134)sin(-158°)

The sum of torques in equilibrium (α =0) gives

(2.2m)(180N)sin(141°) + (1.65m)(F₂)sin(103°) + (1.1m)(134N)sin(-158°) = 0

F₂ ≈ 121 N.

b) Just use Newton's 2nd Law in the x-direction. It seems odd but Newton's 2nd Law doesn't specify where forces need to be on an object so although the forces are at different parts of the bar you can add them. Let's assume the horizontal component is to the right.

F_x - T₁cos(29°) + T₂cos(35°) = 0

F_x ≈ -42 N

c) Similarly

F_y - T₁sin(29°) - T₂sin(35°) - W = 0

F_x ≈ 296 N

Double check the algebra