Question

7.19:

In example 7.5 (page 258) the two accelerations are given by equations 7.66 and 7.67. Check that the acceleration of the block is given correctly in the limit M --> 0. [You need to find the components of this acceleration relative to the table]

Figure 7.8 A block of mass m slides down a wedge of mass M, which is free to slide over the horizontal table.

I am mostly just confused on what it means to find the components relative to the table?

Answer 1

MONTH DATE Up Pg Dn 2 DOO So what happens when M o . In this limit sunely the center of mass (CM) shits completely inside the CM of the mass m. So what are the condinates along the table of the mass m te the c.l of the system. It is 92 +9, cosa. Cordinate of my Now the only fonce acting on the system is gravibadional force la to the table. In this situation the condinate of the C.M shouldn't move along x direction (ie along the table]. That means in the M o . limit the object with mass in where the C.M is situated shouldn't move along se axis. To check the Mto limit of the formula we. have to check whether 42+2, cosa is constant with time or not.. The given formulas are AA1 End - ä cosa Mum maceta Mm HELIOS 3 In the limit M o - 4 casa & ;= a/sing Taking the boundary condition that at to d o we get. =d.cod . since Thus, $(4+4, casa) – 4,14, cosa =) pos 14 canxo no MIMO Thus component of cordinate of m along the toble doesn't change with E Thus the expression for accelendien is om in Mo Imt. Plz. give a gify the my answer if u like my answer