We read your photo as:

The moment of a force (also known as torque) about a point is calculated by the formula:

Physics

ScanSolve AI Answer

Answer

A) 25 Nm\text{A) } 25 \text{ Nm}

Step-by-step solution

3 steps

The moment of a force (also known as torque) about a point is calculated by the formula:

M=F⋅d⊥M = F \cdot d_{\perp}

where MM is the moment, FF is the magnitude of the force, and d⊥d_{\perp} is the perpendicular distance from the pivot point to the line of action of the force.

Alternatively, if the force FF is applied at a distance rr from the pivot and makes an angle θ\theta with the lever arm, the moment can be calculated as:

M=F⋅r⋅sin⁡(θ)M = F \cdot r \cdot \sin(\theta)

Here, θ\theta is the angle between the force vector and the position vector (lever arm) from the pivot to the point of application of the force.

  1. Identify the given values from the diagram.

    • Magnitude of the force, F=10 NF = 10 \text{ N}.
    • Length of the lever arm (distance from O to the point of application), r=5 mr = 5 \text{ m}.
    • The angle shown is 30∘30^\circ between the force vector and the horizontal bar.

  2. Determine the effective angle for the moment calculation.

    The angle θ\theta in the formula M=F⋅r⋅sin⁡(θ)M = F \cdot r \cdot \sin(\theta) is the angle between the force vector and the lever arm. In this case, the lever arm is horizontal. The force is directed downwards and to the left, making an angle of 30∘30^\circ with the horizontal.
    The component of the force perpendicular to the lever arm is F⊥=Fsin⁡(30∘)F_{\perp} = F \sin(30^\circ).

  3. Calculate the moment.

    Using the perpendicular component of the force:

    M=F⊥⋅rM = F_{\perp} \cdot r M=(Fsin⁡(30∘))⋅rM = (F \sin(30^\circ)) \cdot r

    Substitute the values:

    M=(10 N⋅sin⁡(30∘))⋅5 mM = (10 \text{ N} \cdot \sin(30^\circ)) \cdot 5 \text{ m}

    We know that sin⁡(30∘)=0.5\sin(30^\circ) = 0.5.

    M=(10 N⋅0.5)⋅5 mM = (10 \text{ N} \cdot 0.5) \cdot 5 \text{ m} M=5 N⋅5 mM = 5 \text{ N} \cdot 5 \text{ m} M=25 NmM = 25 \text{ Nm}

    The moment of the 10 N force about O is 25 Nm.

    The final answer is A) 25 Nm\text{A) } 25 \text{ Nm}.

Was this helpful?

Try one yourself

A new problem that tests the same skill.

Still stuck on this one?

Ask a follow-up, and the answer starts from this question.

Got a different question?Ask your own question

More Physics Questions

QThe moment of a force about a point (also called torque) is calculated as the product of the force…QWhat is moment of a forceQQuestion 2 (a): What is meant by moment of a force?QCalculate the moment of force F about the axis AC, given the coordinates of A, B, C and the force…QCompute the moment MO of the 1.2-kN force about the axis O-O.
Browse by topic:Mechanics
See all Physics questions →
Got a syllabus? Photograph it and I'll tell you what's due and what's worth the most.