nstead of the block being pulled parallet with a force of, FT, it will be pulled

Question

nstead of the block being pulled parallet with a force of, FT, it will be pulled at an angle of theta below the horizontal. Given the following information find the angle at which the block will tip? The mass of the block is 200g, the coeffcient of friction is .45, the lenght and height of the block, respectavly are 5.2cm and 8.1 cm with the following range of FT solve when theta is 10N, 5N, 1.96N

Fslide= usMg Ftip=(L/2H)Mg us=tan(theta)

i rotation axis Mg Figure 2: Free-Body Diagram for Tipping point lies on the chosen axis of rotation. In other words, the normal force acts at the right edge of the block instead of in the middle. That this is the case can be easily understood by asking where the normal force would act if Fr were increased any further. If that were to happen, the block would ti over so that its right edge would be the only point of contact with the table Thus, just before tipping, N must be acting at the right edge of the block

Explanation / Answer

Because the cabinet has uniform density overall, the center of gravity will be equidistant from the top and bottom, and from the left and right.
x = 5.2 / 2 = 2.6 cm
y = 8.1 / 2 = 4.05 cm

The tipping point is when the point calculated above is exactly over the edge that contacts the ground.

Draw a picture. The x and y values from above are the legs of a right triangle. When balanced the hypotenuse will be vertical. In this orientation, + q = 90, where q is the angle between the base of the cabinet and the hypotenuse.

tan(q) = y/x = 4.05/2.6=1.55
q = arctan( 1.55) = 57.2
+ 57.2 = 90
=32.8

This. Is angle of tip value

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