Example 13  J-E-L-L-O

A block of Jell-O is resting on a plate. Figure 10.32a gives the dimensions of the block. You are bored, impatiently waiting for dinner, and push tangentially across the top surface with a force of F=0.45 N, as in part b of the drawing. The top surface moves a distance DX= 6.0×10–3 m relative to the bottom surface. Use this idle gesture to measure the shear modulus of Jell-O.

(a) A block of Jell-O and (b) a shearing force applied to it.
Figure 10.32  (a) A block of Jell-O and (b) a shearing force applied to it.

Reasoning  The finger applies a force that is parallel to the top surface of the Jell-O block. The shape of the block changes, because the top surface moves a distance DX relative to the bottom surface. The magnitude of the force required to produce this change in shape is given by Equation 10.18 as . We know the values for all the variables in this relation except S, which can be determined.

Solution Solving Equation 10.18 for the shear modulus S, we find that , where A=(0.070 m)(0.070 m) is the area of the top surface, and L0=0.030 m is the thickness of the block:

Jell-O can be deformed easily, so its shear modulus is significantly less than that of a more rigid material like steel (see Table 10.2).



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