Engineering Mechanics Test 2

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Engineering Mechanics Test 2

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Question 1
1 / -0

A block of mass 200 gm moves with uniform speed in a circular groove. With vertical side walls of radius 35 cm. Find the normal contact force by the side wall, if the block takes 3 seconds to complete one round.

A
0.198 N

B
0.211 N

C
0.307 N

D
0.420 N

Solution

Concept:
Speed of block, V = 2πr/t
Normal force = m × a_n, where a_n = v²/r and m = mass of the block
Calculation:
Given:
mass (m) = 200 gm, radius of the side walls (r) = 35 cm, time (t) = 3 sec
v = 2πr/t = 0.733 m/s
a_n = v²/r = 1.535 m/s²
N = m × a_n
∴ N = 0.307 N
Question 2
1 / -0

Two forces \({\vec F_1} = \hat i + 2\hat y + 3\hat x\) and \({\vec F_2} = 4\hat i + 3\hat y - \hat x\) are acting on a rigid body. The component of \({\vec F_2}\) in direction of \({\vec F_1}\) is ______ N

A
1.5-2

B
1.54-21

C
1.57-26

D
1.59-22

Solution

Concept:
Component of \({\vec F_2}\) in \({\vec F_1} = {\vec F_2}\cos \theta \)
\({\rm{Where}},\cos \theta = \frac{{{{\vec F}_1}.{{\vec F}_2}}}{{\left| {{{\vec F}_1}} \right|\left| {{{\vec F}_2}} \right|}}\)
Calculation:
\({\vec F_1}.{\vec F_2} = 4 + 6 + \left( { - 3} \right) = 7\)
\(\left| {{{\vec F}_1}} \right| = 3.74\;N;\left| {{{\vec F}_2}} \right| = 5.099\;N\)
cos θ = 0.367
∴ F₂ cos θ = 5.099 × 0.367
∴ F₂ cos θ = 1.87 N
Question 3
1 / -0

A flywheel of mass 5000 kg and having a radius of gyration 1 m, looses its speed form 400 rpm to 280 rpm in 2 minutes Determine the retarding torque acting on flywheel.

A
1222 N.m

B
1745 N.m

C
524 N.m

D
31425 N.m

Solution

Explanation:
Initial angular velocity (ω_i):
\({ω _o} = \frac{{2\pi {N_o}}}{{60}} = \frac{{2\pi \times 400}}{{60}} = 41.89\;rad/s\)
Final angular velocity (ω_f):
\({ω _i} = \frac{{2\pi {N_o}}}{{60}} = \frac{{2\pi \times 280}}{{60}} = 29.32\;rad/s\)
I = mk² = 5000 × (1)² = 5000 kg – m²
Angular acceleration:
ω_f = ω_i + αt
\(\alpha = \frac{{{ω _f} - {ω _i}}}{t} = - \frac{{41.89 - 29.32}}{{2 \times 60}} = - 0.10475rad/{s^2}\)
Now,
Retarding torque acting on the flywheel
∴ T = Iα
∴ T = 5000 × 0.10475
∴ T = 523.75 N.m
Question 4
1 / -0

A boat of mass 300 kg moves according to the equation x = 1.2t² - 0.2t³. When the force will become zero?

A
2 s

B
1 s

C
6 s

D
2.8 s

Solution

Explanation:
From Newton's second law of motion, the force acting on a body is given by
F = ma
If force is zero then its acceleration must be zero
F = 0 ⇒ a = 0
\(v=\frac{{{d}}x}{d{{t}}}\)
\(a=\frac{{{d}}v}{d{{t}}}= \frac{{{d}{^2}}x}{d{{t^{2}}}}\)
Now,
If a = 0,then \(\frac{{{d}^{2}}x}{d{{t}^{2}}}=0\)
x = 1.2t² - 0.2t³
\(\frac{dx}{dt}=2.4t-0.6{{t}^{2}}\)
\(\frac{{{d}^{2}}x}{d{{t}^{2}}}=2.4-1.2~t=0\)
t = 2.4/1.2
∴ t = 2 s