Question

Does anyone hereknow how to solve question 2.18 (the last question) from http://ocw.mit.edu/courses/civil-and-environmental-engineering/1-050-solid-mechanics-fall-2004/readings/emech2_04.pdf
If possible please draw me a free body diagram.

Answer 1

Friction is what is going to hold the block in place against the force of gravity. The left tong on the vice grip will have a force component pushing the block to the right and the right tong will have a force component pushing the block to the left. These forces are the normal force which establish a force of static friction between the block and the grip. The weight of the block will try to make the block fall downwards. It appears as thought the applied force on the vice grip points upward with a magnitude =W. This is the "opposite" side on a right angle triangle with angle theta in the diagram. We want the "adjacent" side, which is our horizontal component. The force component on each side of the block (which compresses the block) is therefore:\[F_{horizontal}=\frac{W}{\tan \theta}\]One points left and the other points right so the total vectorial sum of the forces is zero (which makes sense since the block doesn't accelerate left or right). The total magnitude of the compression force is F=2W/tan(theta).

Answer 2

Friction is computed by:\[F_{friction}=\mu_{s} N\]Here the normal force, N is of magnitude 2W/tan(theta) as shown above. The frictional force acts upwards to counteract the force of gravity and hold the block in place. So, the magnitude of the frictional force must be equal to W. We want the static friction coefficient. \[\mu _{s}=\frac{W \tan \theta}{2W}\]So,\[\mu_{s}=\frac{1}{2}\tan 30=\frac{\sqrt{3}}{6}\]

Answer 3

thank you so much~ your explaination was very clear!! I guess i was over thinking. Thank you for your help. I appreciate it. ^^

Answer 4

no problem. I hope it is correct :)