How is this done?(physics/laws of motion)

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1 Answer
Gió
Jan 28, 2016

I think a) #15kg#

Explanation:

Let us consider all the forces involved:
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Where:
#W="weight"=mg#
#f="friction"=mu_sN#
#N="normal reaction"#;
#T=#tension.
and apply Newton`s Second law to the two blocks, CA and B with the condition that they must stay at rest (#a=0#).
We get:
for CA (along horizontal and vertical):
#T-f=0#
#N-W_x=0#
for B:
#T-W_B=0#

Let us expand a bit the three expressions:
#T-mu_sN=0#
#N=(m_c+m_a)g#
#T=m_Bg#
Let us use the fact that we have the same tension and substitute the third equation into the first and use the expression for #N# to get:
#m_Bcancel(g)-mu_s(m_c+m_a)cancel(g)=0#
and with our data:
#5-0.2(10+m_c)=0#
so #m_c=15kg#

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