Figure below shows a block of mass 35 kg
resting on a table. The table is so rough that
it offers a self adjusting resistive force 10%
of the weight of the block for its sliding
motion along the table. A 20 kg wt load is
attached to the block and is passed over a
pulley to hang freely on the left side. On the
right side there is a 2 kg wt pan attached to
the block and hung freely. Weights of 1 kg
wt each, can be added to the pan. Minimum
how many and maximum how many such
weights can be added into the pan so that
the block does not slide along the table
Answers
Answer:
So it requires minimum 15 and maximum 22 such weights that we can add into the pan
Explanation:
As we know that the weight of the block placed on the table is
now the resistive force due to friction is 10% of the weight so we have
Now weight suspended on the left side is given as
Now in order to make the system at equilibrium we can make force balance equation as
so total weight on the right side is 165 N
Now we know that the weight of the pan is 2 kg so extra weight required is
so for nearest integer we need 15 weights
Now similarly for maximum number we have force balance as
So we will have
for maximum number of weight we need 22 such masses
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Topic : Force Balance
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Explanation:
Figure below shows a block of mass 35 kg
resting on a table. The table is so rough that
it offers a self adjusting resistive force 10%
of the weight of the block for its sliding
motion along the table. A 20 kg wt load is
attached to the block and is passed over a
pulley to hang freely on the left side. On the
right side there is a 2 kg wt pan attached to
the block and hung freely. Weights of 1 kg
wt each, can be added to the pan. Minimum
how many and maximum how many such
weights can be added into the pan so that
the block does not slide along the table