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# PHYSICS PLEASE HELP! PARTS B, C, D, AND E!?

A rod is attached to a vertical beam by a hinge at point A and a horizontal cable at point B, as shown in the figure. The angle between the rod and the beam is θ and the distance between points A and B is L.

B) Let the tension in the cable be T = Ti. Enter a vector expression for the torque T1 on the rod about the hinge due to that tension, in terms of T, L, θ, and the unit vectors i, j, k.

C) Given the values T = 399 N, L = 9.1 m, and θ = 21°, calculate the magnitude of the torque T1, in newton meters.

D) A force F = -Fj acts on the rod at point C, whose distance from the hinge is fL, 0<f<1. Refer to the figure. Enter a vector expression for the torque T2 on the rod about the hinge due to the force F, in terms of F, L, f, θ, and the unit vectors i, j, k.

E) For the values L = 9.1 m, θ= 21°, T = 399 N, and f = 0.84, calculate the magnitude F of the force F, in newtons, needed to balance the torque from the tension in the cable.

### 2 Answers

- Steve4PhysicsLv 71 year agoFavorite Answer
We assume the rod’s weight is negligible.

With the usual right-hand convention the +z direction is ‘out of’ the diagram: https://i.stack.imgur.com/9KgUe.png

Here’s a detailed explanation.

_____________________________________

A)

Long method

If we take A as the origin, B has coordinates (-Lsinθ, Lcosθ, 0).

So the position vector of B with respect to A is r = (-Lsinθi + Lcosθj + 0k).

Tension pulling B to right is (Ti + 0j + 0k).

Torque is:

τ₁ = rXT (cross product)

=

|i. . . . . j. . . . . k|

|-Lsinθ Lcosθ 0|

|T. . . . 0. . . . .0|

= -LTcosθ k

You might need to enter it as 0i + 0j -LTcosθk.

The shorter method is:

The lever arm length is Lcosθ so the magnitude of the torque is LTcos θ .

Using the right hand grip rule, the direction of the torque is ‘into’ the diagram, i.e. the -z direction. So the torque is -LTcosθ k.

________________________

C)

|τ₁| = 9.1*399*cos(21°) = 3390 Nm to 3 sig. figs,

________________________

D)

Point C has coordinates (fLsinθ, fLcosθ, 0) and F acts in the -y direction. So you can use the long method from part A). But using the short method:

The lever arm length is fLsinθ so the magnitude of the torque is fLFsinθ .

Using the right hand grip rule, the direction of the torque is ‘into’ the diagram, I,.e. the -z direction. So the torque is τ₂ = -fLFsinθ.

________________________

E)

In equilibrium τ₁+τ₂ =0 (so |τ₁|=|τ₂|) which gives:

fLFsinθ = 3390

F = 3390/(fLsinθ)

= 3390/(0.84*9.1*sin(21°))

= 1238N which you may need to round 2 or 3 sig. figs.

Check my arithmetic though.