GATE Mechanical (ME) Engineering Mechanics Free Online Test 2026

Option C is correct: 2.10 kN (C).

Consider equilibrium at joint E. The member EB is inclined at 45° (from E to B: 2 m horizontally and 2 m vertically). Sum of horizontal forces at E gives

F_EB cos45° = 900 N, so F_EB = 900·√2 = 1.273 kN (T).

The vertical component of this force at E is F_EB·sin45° = 900 N (downward).

Now consider equilibrium at joint D. The member DB has horizontal span 2 m and vertical drop 4 m, so for DB

cosθ = 2/√20 = 0.4472136. From horizontal equilibrium at D,

F_DB cosθ = 600 N, hence F_DB = 600 / 0.4472136 = 1.342 kN (T).

The vertical component of DB at D is F_DB·(4/√20) = 1.342·0.894427 = 1200 N (downward).

Vertical equilibrium at D therefore requires an upward 1200 N from ED; this means ED = 1.200 kN (compression) (it pushes down on E by 1200 N).

At joint E the downward forces are the vertical component of EB (900 N down) and the force from ED on E (1200 N down), totalling 2100 N down. To balance these, EA must push up on E with 2100 N. A member that pushes on the joint is in compression.

Therefore EA = 2.10 kN (compression). The correct option is Option C.

Joint A:-

Joint C:-

• The weight 'W' is supported by two cables, one of which makes an angle θ with the horizontal.
• To minimize the tension in this cable, consider the equilibrium of forces.
• The vertical component of the tension must equal the weight (T sin θ = W).
• For minimal tension, T = W/sin θ should be minimized.
• The function 1/sin θ is minimized when θ = 30°, which maximizes sin θ.

To find the acceleration of a passenger on a carnival ride rotating in a vertical circle, we can use the formula for centripetal acceleration:

• Centripetal acceleration is given by the formula: a = v²/r, where:
  • v = linear speed of the passenger
  • r = radius of the circular path
• The ride completes one full circle in 2 seconds. Therefore, the time period T = 2 seconds.
• We can calculate the linear speed v using the formula: v = 2πr/T.
• Substituting the values:
  • r = 2 m
  • T = 2 s
  • v = 2π(2)/2 = 2π m/s
• Now substituting v back into the centripetal acceleration formula:
  • a = (2π)² / 2
  • a = 4π² / 2 = 2π²
  • Using π² ≈ 9.87, we get:
  • a ≈ 19.74 m/s²

Thus, the acceleration of the passenger is approximately 19.7 m/s².

Option C. The normal force at the contact point is N ≈ 35.4 N, so the closest option is 35 N.

Step 1: Weight of the ladder: W = 10 × 9.81 = 98.1 N.

Step 2: Let the horizontal distance from the foot A to the sphere centre O be h, and the distance along the ladder from A to contact point B be s. Projection of AO on the ladder gives s = 0.5h + (5√3)/2. The perpendicular distance from O to the ladder must equal the radius R = 5 m, which leads to h = s. Solving gives s = h = 5√3 = 8.660254 m.

Step 3: Position of the weight (midpoint of ladder, at 6.25 m from A) projected on horizontal: x_w = 6.25 cos60° = 3.125 m.

Step 4: Coordinates of contact B: x_b = s cos60° = 4.330127 m, y_b = s sin60° = 7.5 m.

Step 5 (moments about A): Taking moments about A, the moment of the normal N (acting perpendicular to the ladder) isN [0.5 x_b + (√3/2) y_b] and must balance the moment of the weight W x_w. Thus

N = (W x_w) / [0.5 x_b + (√3/2) y_b].

Numerical substitution: denominator = 0.5×4.330127 + 0.8660254×7.5 = 8.660254. Hence

N = 98.1 × 3.125 / 8.660254 ≈ 35.4 N.

Conclusion: The sphere exerts a normal force of about 35.4 N on the ladder, so option C (35 N) is correct.