A uniform disk of mass m = 2.4 kg and radius R = 25 cm can rotate about an axle through its center. Four forces are acting on it as shown in the figure. Their magnitudes are F1 = 8.5 N, F2 = 1.5 N, F3 = 6.5 N and F4 = 6.5 N. F2 and F4 act a distance d = 3.5 cm from the center of mass. These forces are all in the plane of the disk. Write an expression of the magnitude of the torque due to the force F3

Newton's second law for rotational motion allows to find the results for the angular acceleration of the disk and the torque of force 3 are:

Given parameters

To find

Newton's second law for rotational motion states that torque is equal to the product of the moment of inertia times the angular acceleration

             ∑τ = I α

             τ = F x r

Where the bold letters indicate vectors, τ is the torque, I the moment of inertia, α the angular acceleration, F the force and r the distance.

Torque is the vector magnitude has modulus and direction, the modulus can be found by developing the vector product

        τ  = F r sin θ

In general, for rotational movements, the reference system is placed in the turning point, in this case the axis of the disk and the positive direction is for counterclockwise turns, let's apply to this problem

   Σ τ = F₁ R - F₂ d + F₃ 0 + F₄ d sin 53

   Σ τ = 8.5 0.25 - 1.5 3.5 + 0 + 6.5 3.5 sin 53

   Σ τ = 15.0 N m

As there is a net torque the disk is being accelerated, let's look for this angular acceleration

The moment of inertia of a disk is

        I = ½ m r²

        I = ½ 2.4 0.25²

        I = 0.075 Kg m²

         α =  

         α = 15 / 0.075

         α = 200 rad / s²

The expression for the torque of force 3 is

    τ₃ = F₃ d₃ sin 45

    τ₃ = F₃ sin 45 0

    τ₃ = 0

In conclusion, using Newton's second law for rotational motion, we can find the angular acceleration of the disk and the torque of the force  3 are:

       α = 200 rad / s²

      τ₃ = F₃ d₃ sin 45 = 0

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weight of the box is given as

here this is placed on the inclined of angle 42 degree

now this weight will have two components one is along the incline other is perpendicular to it

so here we need to apply a force parallel to incline opposite to its component of weight

so here the component of weight is given as

here we have

so it requires 75 n force to move it upwards