**Problem Description:**In the figure shown, the 120-lb force is along line \(OA\), the 200-lb force is perpendicular to line \(OA\), the 260-lb force is along \(x\)-direction, and the 350-lb force is at a 25° angle with \(x\)-direction. Replace the loading on the triangular plate by an equivalent resultant force (indicate the magnitude and direction of the force) and a couple moment (indicate the magnitude and direction of the moment) acting at point \(O\).**Diagram Instructions:**- The diagram is a triangular plate with: - A base of 4 ft along the \(x\)-direction. - A height of 3 ft along the \(y\)-direction. - A force of 120 lb directed along line \(OA\), which makes a 45° angle with the horizontal base. - A force of 200 lb perpendicular to line \(OA\), directed upwards along the \(y\)-direction. - A force of 260 lb directed horizontally to the left along the \(x\)-direction. - A 350-lb force making a 25° angle with the horizontal at point \(A\), located 3 ft above the origin (\(O\)).The notation on the diagram includes:- Origin \(O\) at the base left corner of the triangle.- Point \(A\) is located 3 ft directly above \(O\) on the y-axis.- The \(y\)-axis is vertical and the \(x\)-axis is horizontal.- A 600 lb-ft couple moment (\(M\)) acting clockwise at point \(A\).The task is to:1. Resolve all the forces into their components along the x and y directions.2. Sum up the force components to find the equivalent resultant force.3. Calculate the moments created by the forces about point \(O\).4. Sum the moments to find the equivalent couple moment at \(O\).**Step-by-Step Solution:**1. **Resolve Forces into Components:** - 120-lb force along \(OA\): - \(F_{x1} = 120 \cos(45^\circ) = 84.85 \text{ lb}\) - \(F_{y1} = 120 \sin(45^\circ) = 84

Answered: Image /qna-images/answer/77f01a8c-bb05-410a-9e17-5e271e5a8a10.jpg

In the figure shown, the 120-lb force is along line OA, the 200-lb force is perpendicular to line OA, the 260-lb force is along x-direction, and the 350-lb force is at a 25° angle with x-direction. Replace the loading on the triangular plate by an equivalent resultant force (indicat the magnitude and direction of the force) and a couple moment (indicate the magnitude and direction of the moment) acting at point O. 600 lb ft 350 lb y A 250 200 lb 3 ft 2.5 ft 260 lb 120 lb 4 ft

In the figure shown, the 120-lb force is along line OA, the 200-lb force is perpendicular to line OA, the 260-lb force is along x-direction, and the 350-lb force is at a 25° angle with x-direction. Replace the loading on the triangular plate by an equivalent resultant force (indicat the magnitude and direction of the force) and a couple moment (indicate the magnitude and direction of the moment) acting at point O. 600 lb ft 350 lb y A 250 200 lb 3 ft 2.5 ft 260 lb 120 lb 4 ft

Elements Of Electromagnetics

7th Edition

ISBN:9780190698614

Author:Sadiku, Matthew N. O.

Publisher:Sadiku, Matthew N. O.

ChapterMA: Math Assessment

Section: Chapter Questions

Problem 1.1MA

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Concept explainers

Clutches, Brakes, Couplings, and Flywheels

The clutch is described as a machine element used to connect or disconnect two shafts together to transmit torque or power. The clutch works on the principles of friction. The clutch is commonly used in automobiles to transfer power from the driving shaft to the driven shaft.

Question

**Problem Description:**

In the figure shown, the 120-lb force is along line \(OA\), the 200-lb force is perpendicular to line \(OA\), the 260-lb force is along \(x\)-direction, and the 350-lb force is at a 25° angle with \(x\)-direction. Replace the loading on the triangular plate by an equivalent resultant force (indicate the magnitude and direction of the force) and a couple moment (indicate the magnitude and direction of the moment) acting at point \(O\).

**Diagram Instructions:**

- The diagram is a triangular plate with:
- A base of 4 ft along the \(x\)-direction.
- A height of 3 ft along the \(y\)-direction.
- A force of 120 lb directed along line \(OA\), which makes a 45° angle with the horizontal base.
- A force of 200 lb perpendicular to line \(OA\), directed upwards along the \(y\)-direction.
- A force of 260 lb directed horizontally to the left along the \(x\)-direction.
- A 350-lb force making a 25° angle with the horizontal at point \(A\), located 3 ft above the origin (\(O\)).

The notation on the diagram includes:
- Origin \(O\) at the base left corner of the triangle.
- Point \(A\) is located 3 ft directly above \(O\) on the y-axis.
- The \(y\)-axis is vertical and the \(x\)-axis is horizontal.
- A 600 lb-ft couple moment (\(M\)) acting clockwise at point \(A\).

The task is to:
1. Resolve all the forces into their components along the x and y directions.
2. Sum up the force components to find the equivalent resultant force.
3. Calculate the moments created by the forces about point \(O\).
4. Sum the moments to find the equivalent couple moment at \(O\).

**Step-by-Step Solution:**

1. **Resolve Forces into Components:**
- 120-lb force along \(OA\):
- \(F_{x1} = 120 \cos(45^\circ) = 84.85 \text{ lb}\)
- \(F_{y1} = 120 \sin(45^\circ) = 84

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