Assume the three blocks ( m 1 = 1.0 kg, m 2 = 2.0 kg, and m 3 = 3.5 kg) portraye

Question

Assume the three blocks (m1 = 1.0 kg, m2 = 2.0 kg, and m3 = 3.5 kg) portrayed in the figure below move on a frictionless surface and a force F = 36 N acts as shown on the 3.5-kg block.

Assume the three blocks (m1 = 1.0 kg, m2 = 2.0 kg, and m3 = 3.5 kg) portrayed in the figure below move on a frictionless surface and a force F = 36 N acts as shown on the 3.5-kg block. Determine the acceleration given this system. m/s2 (to the right) Determine the tension in the cord connecting the 3.5-kg and the 1.0-kg blocks. N Determine the force exerted by the 1.0-kg block on the 2.0-kg block. N

Explanation / Answer

(a) since the surface is frictionless, only external force that acts on the system is F.

so, acceleration of system=force on the system/mass of the system=36/(1+2+3.5) m/s^2=5.53 m/s^2

the direction of acceleration is same of that of force, i.e towards +ve x(increasing right side).

(b)[let T1 be the tension that acts on the box of 1kg(in right direction) & T2 be be the tension that acts on m(towards left)

tension in the cord is same everywhere, assuming it's massless, if it was not so, say T1>T2
then the resulting tension would act on the cord and accelerate it, i.e T1-T2=mass of cord*some acceleration

since mass of cord=0, T1-T2=0=>T1=T2=T(say)]

There will be contact force P working between 1kg & 2kg boxes

for box of 2 kg, P(contact force exerted by 1kg box)=2*5.53 N
thus P=11.06 N

for box of mass 1kg,
T-P(contact force exerted by 2kg box on 1kg box)=1*5.53
T=P+5.53 N=16.59 N

(c) already determined, P=11.06 N

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