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Homework Statement
Fig. 4.3
A 5.0-kg block and a 4.0-kg block are connected by a 0.6-kg rod. The links between the blocks and the rod are denoted by A and B. A force F is applied to the upper block.In Fig. 4.3, the force in link B is 40 N. The acceleration of the blocks and rod assembly, including direction, is closest to:
1) zero
2) 2.4 m/s2, downward
3) 1.2 m/s2, downward
4) 2.4 m/s2, upward
5) 1.2 m/s2, upward
Homework Equations
we have to use Newton's Laws of Motion to find the answer.
The Attempt at a Solution
I tried to figure out the acceleration for each of 5.0-kg block, 4.0-kg block and 0.6-kg rod using these equations:
for the 5.0-kg block \ ƩF=ma ==> F - m1g - Fa = m1a ==> F-Fa = m1(a+g)
for 4.0-kg block \ ƩF=ma ==> Fb - m2g = m2a ==> Fb = m2(a+g)
for the 0.6-kg rod \ Fa + Fb ==> (F(m1+m2+m3) - m1F)/(m1+m2+m3) + (m2F)/(m1+m2+m3)and i could do that for the 4.0-kg block and 0.6-kg rod but i couldn't with the 5.0-kg block because F is unknown for me!- I find Fa and Fb this way :
First ƩF for the hole system is ƩF=ma ==> F-(m1+m2+m3)g = (m1+m2+m3)a
so, a = F/(m1+m2+m3) -g
for m1 ƩF=ma ==> F-Fa-m1g = m1a
so, F-Fa = m(a+g) = m1F/(m1+m2+m3) ==> Fa = F(m1+m2+m3)-m1F/(m1+m2+m3) ==> Fa = 0.48N
and Fb = 40N (Given in the question)
m1 = mass of 5.0-kg block
m2 = mass of 0.6-kg rod
m3 = mass of 4.0-kg block
Fa = Force of A
Fb = Force of B
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