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3 parts
a

41.0 Nm\text{41.0 Nm}

b

20.1 Nm\text{20.1 Nm}

c

0 Nm\text{0 Nm}

Step-by-step solution

6 steps

To solve this problem, we will use the formula for the moment of a force, which is given by M=F×d×sin⁡(θ)M = F \times d \times \sin(\theta). Here, FF is the magnitude of the force, dd is the perpendicular distance from the pivot point to the line of action of the force (or the distance from the pivot to the point where the force is applied), and θ\theta is the angle between the force vector and the position vector from the pivot to the point of application.

Part a· 2 steps

Determine the moment of W about Q

  1. Identify the force, distance, and angle.

    The weight force W=20W = 20 N. This force acts downwards at the center of the bar PQ. The total length of the bar PQ is 5 m, so the center is at 2.5 m from either end. Therefore, the distance dd from point Q (the pivot) to the point where W acts is 2.5 m. The bar PQ makes an angle of 35° with the horizontal. Since the weight acts vertically downwards, the angle θ\theta between the bar (from Q to the midpoint) and the vertical direction of the weight is 90∘−35∘=55∘90^\circ - 35^\circ = 55^\circ.

  2. Apply the moment formula.

    MW=W×d×sin⁡(θ)M_W = W \times d \times \sin(\theta)

    Substitute the values:

    MW=20 N×2.5 m×sin⁡(55∘)M_W = 20 \text{ N} \times 2.5 \text{ m} \times \sin(55^\circ)

    Calculate the product:

    MW=50×0.81915M_W = 50 \times 0.81915 MW=40.9575 NmM_W = 40.9575 \text{ Nm}

    Rounding to three significant figures:

    MW=41.0 NmM_W = \boxed{\text{41.0 Nm}}

Part b

20.1 Nm\text{20.1 Nm}

Determine the moment of FQF_Q about P

  1. Identify the force, distance, and angle.

    The force FQ=7F_Q = 7 N. This force is applied at point Q. The pivot point is P. The distance dd from P to Q is the full length of the bar, which is 5 m. The force FQF_Q acts horizontally. The bar PQ makes an angle of 35° with the horizontal. Therefore, the angle θ\theta between the bar (from P to Q) and the horizontal direction of FQF_Q is 35∘35^\circ.

  2. Apply the moment formula.

    MFQ=FQ×d×sin⁡(θ)M_{F_Q} = F_Q \times d \times \sin(\theta)

    Substitute the values:

    MFQ=7 N×5 m×sin⁡(35∘)M_{F_Q} = 7 \text{ N} \times 5 \text{ m} \times \sin(35^\circ)

    Calculate the product:

    MFQ=35×0.57358M_{F_Q} = 35 \times 0.57358 MFQ=20.0753 NmM_{F_Q} = 20.0753 \text{ Nm}

    Rounding to three significant figures:

    MFQ=20.1 NmM_{F_Q} = \boxed{\text{20.1 Nm}}

Part c

0 Nm\text{0 Nm}

Determine the moment of FPF_P about P

  1. Identify the force and its point of application relative to the pivot.

    The force FPF_P is applied at point P. The pivot point is also P.

  2. Apply the moment formula.

    When a force acts directly at the pivot point, the distance dd from the pivot to the point of application of the force is zero. Therefore, the moment created by this force about that pivot is zero.

    MFP=FP×0M_{F_P} = F_P \times 0 MFP=0 NmM_{F_P} = \boxed{\text{0 Nm}}
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