EIT - Engineer In Training Statics Questions and Answers

Question 1

A 10 kg block rests on a plane inclined at 30 degrees. The coefficient of static friction between the block and the plane is 0.30. What is the minimum horizontal force (P) required to prevent the block from sliding down the plane? (Use g = 9.81 m/s²)

Accepted Answer

12.4 N

Answer

First, draw a free-body diagram. The forces acting on the block are its weight (W), the normal force (N), the friction force (F_f), and the applied horizontal force (P). For minimum P, the friction force will act up the incline, opposing the impending motion downwards. Sum forces perpendicular to the incline: ΣF_y = N - Wcos(30) - Psin(30) = 0. Sum forces parallel to the incline: ΣF_x = F_f + Pcos(30) - Wsin(30) = 0. We know W = mg = 10 kg * 9.81 m/s² = 98.1 N. The friction force at impending motion is F_f = μ_s * N. Substitute N from the first equation into the friction formula: F_f = μ_s * (Wcos(30) + Psin(30)). Now substitute this into the second equation: μ_s * (Wcos(30) + Psin(30)) + Pcos(30) - Wsin(30) = 0. Rearrange to solve for P: P * (μ_s*sin(30) + cos(30)) = Wsin(30) - μ_s*Wcos(30). P = W * (sin(30) - μ_s*cos(30)) / (μ_s*sin(30) + cos(30)). P = 98.1 * (0.5 - 0.3*0.866) / (0.3*0.5 + 0.866) = 98.1 * (0.2402) / (1.016) ≈ 12.4 N.

Question 2

A simple pin-jointed truss is loaded as shown. What is the force in member AB? (A positive value indicates tension, a negative value indicates compression). The truss has a 1000 N vertical load at joint B. Joint A is a pin support and Joint C is a roller support. The angle at Joint A and Joint C is 45 degrees.

Accepted Answer

-707 N

Answer

We can use the Method of Joints, starting at joint B. Draw the free-body diagram for joint B. The forces acting on it are the external 1000 N load (downwards), the force from member AB (F_AB), and the force from member BC (F_BC). The angle for both members with the horizontal is 45 degrees. Sum of forces in the vertical direction (ΣF_y = 0): F_AB*sin(45°) + F_BC*sin(45°) - 1000 N = 0. Sum of forces in the horizontal direction (ΣF_x = 0): -F_AB*cos(45°) + F_BC*cos(45°) = 0. From the horizontal equation, since cos(45°) is not zero, we find that F_AB = F_BC. Substitute this into the vertical equation: 2 * F_AB * sin(45°) = 1000 N. F_AB = 1000 / (2 * sin(45°)) = 1000 / (2 * 0.707) = 707 N. However, looking at the FBD of joint B, both F_AB and F_BC must have upward vertical components to counteract the 1000N downward force. This means they must both be pushing on the joint, indicating compression. Therefore, the force is -707 N.

Question 3

A beam is subjected to a triangular distributed load as shown, with a maximum intensity of 60 N/m at the fixed support and decreasing to 0 N/m at the free end over a length of 3 m. What is the magnitude of the equivalent resultant force and its location from the fixed support?

Accepted Answer

90 N, located 1.0 m from the support

Answer

The magnitude of the equivalent resultant force for a distributed load is equal to the area under the load diagram. For a triangle, the area is (1/2) * base * height. Magnitude = (1/2) * (3 m) * (60 N/m) = 90 N. The location of this resultant force acts through the centroid of the distributed load area. The centroid of a triangle is located at 1/3 of the base as measured from the tall end (the fixed support). Location = (1/3) * base = (1/3) * 3 m = 1.0 m from the fixed support.

Question 4

A 100 N force is applied at a 30-degree angle to the end of a 2-meter long wrench, as shown. What is the magnitude of the moment created by this force about point A (the pivot point)?

Accepted Answer

100.0 Nm

Answer

The moment (M) is calculated as M = F * d, where d is the perpendicular distance from the pivot point to the line of action of the force. A common way to solve this is to find the component of the force that is perpendicular to the lever arm. The perpendicular component (F_perp) is F * sin(θ). Here, F_perp = 100 N * sin(30°) = 100 N * 0.5 = 50 N. The moment arm is the full length of the wrench, 2 m. So, M = F_perp * r = 50 N * 2 m = 100 Nm. Alternatively, M = F * r * sin(θ) = 100 N * 2 m * sin(30°) = 100 Nm.

Question 5

Which of the following statements is a fundamental condition for a rigid body to be in static equilibrium?

Accepted Answer

The net force and the net moment about any point are both zero.

Answer

For a rigid body to be in static equilibrium, two conditions must be met. First, the vector sum of all external forces acting on the body must be zero (ΣF = 0). This ensures translational equilibrium (no acceleration). Second, the vector sum of the moments of all external forces about any arbitrary point must be zero (ΣM = 0). This ensures rotational equilibrium (no angular acceleration). The other options are incorrect: a body can be in equilibrium while moving at a constant velocity (dynamic equilibrium), the net moment must be zero, and the concept of a rigid body is an idealization—real bodies deform but can still be analyzed in equilibrium.

(EIT) Engineer In Training Certification Practice Test

(EIT) Engineer In Training Certification Practice Test

EIT - Engineer In Training Statics Questions and Answers