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

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

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Answer 1 · 2016-01-28T18:01:08

I think a) $15 k g$ $15kg$

Explanation:

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

Let us expand a bit the three expressions:
$T - μ_{s} N = 0$ $T-mu_sN=0$
$N = (m_{c} + m_{a}) g$ $N=(m_c+m_a)g$
$T = m_{B} g$ $T=m_Bg$
Let us use the fact that we have the same tension and substitute the third equation into the first and use the expression for $N$ $N$ 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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How is this done?(physics/laws of motion)

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