Engineering Mechanics - Friction - Discussion
Discussion Forum : Friction - General Questions (Q.No. 1)
1.
The boy at D has a mass of 50 kg, a center of mass at G, and stands on a plank at the position shown. The plank is pin-supported at A and rests on a post at B. Neglecting the weight of the plank and post, determine the magnitude of force P his friend (?) at E must exert in order to pull out the post. Take 
B = 0.3 and C = 0.8.
Discussion:
19 comments Page 1 of 2.
Ramakant jhariya said:
4 years ago
229 is the correct answer.
Chethan said:
5 years ago
Take vertical forces equal to zero, Ay + Ny = 490.5.
Take moment at A, then Ny*3=mg*2, By solving we get N=327.
Friction force = static friction coefficient * normal force.
f = 0.3 * 327 = 98.1;
Now take moment at c(post bottom),we get f(0.3+0.4) = Fcos30 * 0.3
By solving the above equation we get F = 264.115 Newtons.
Take moment at A, then Ny*3=mg*2, By solving we get N=327.
Friction force = static friction coefficient * normal force.
f = 0.3 * 327 = 98.1;
Now take moment at c(post bottom),we get f(0.3+0.4) = Fcos30 * 0.3
By solving the above equation we get F = 264.115 Newtons.
Ajay kumar said:
6 years ago
Please solve it.
Ajay kumar said:
6 years ago
How? Please explain the answer.
Shree said:
7 years ago
Please explain in detail @Sumit manna.
Goku said:
8 years ago
Please explain about C @Sumit Manna.
Debolina said:
8 years ago
Please solve it clearly.
James Oketch said:
9 years ago
Not able to reach the answer. Please explain me in detail.
Sumit manna said:
9 years ago
Force create at A=50g/3 and force in B = 100g/3 verticaly both Friction force is 0.3 * 100g/3 at B.
Moment about C : 0.3 * 100g/3 * (0.3 + 0.4) = Fcos30 * 0.3.
So F= 264N.
Moment about C : 0.3 * 100g/3 * (0.3 + 0.4) = Fcos30 * 0.3.
So F= 264N.
Madara said:
9 years ago
Explain the solving method of this question.
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