Three blocks rest on a frictionless, horizontal table, with m1 = 9 kg and m3 = 16 kg. A horizontal force F = 104 N is applied to block 1, and the acceleration of all three blocks is found to be 3.3 m/s2.
1) Find m2
2)What is the normal force between 2 and 3?
I assume these are in a line.
friction= mu(M1+m2+m3)
netforce= totalmass*acceleration
104-mu(totalmass)=totalmass*acceleration
solve for totalmass, then solve for m2
if block2 is pushing block3,
forceonblock3-mu(m3)=m3*a
solve for forceonblock3
I assume these are in a line.
friction= mu(M1+m2+m3)
netforce= totalmass*acceleration
104-mu(totalmass)=totalmass*acceleration
solve for totalmass, then solve for m2
if block2 is pushing block3,
forceonblock3-mu(m3)=m3*a
solve for forceonblock3
OOOPSsss. I forgot g as part of the weight. every where there is a mu, it should read mu*g*masses
what does u stand for in the above equations?
mu is the coefficent of friction...and it is given as zero, which is a very mythical situation. So you get to forget about friciton. Your teacher is too easy.
so what you're saying is:
forceonblock3-mu(m3)=m3*a
forceonblock3-0(m3)=...
forceonblock3=...
To solve this problem, we can use Newton's second law of motion, which states that the net force acting on an object is equal to the mass of the object multiplied by its acceleration.
1) To find the mass of block 2 (m2), we can use the equation:
F = m2 * a
Where F is the applied force and a is the acceleration.
Plugging in the values we know:
104 N = m2 * 3.3 m/s^2
Dividing both sides by 3.3 m/s^2, we get:
m2 = 104 N / 3.3 m/s^2
m2 = 31.515 kg (rounded to three decimal places)
Therefore, the mass of block 2 is approximately 31.515 kg.
2) To find the normal force between blocks 2 and 3, we need to consider the forces acting on block 2. Since the table is frictionless, the only forces acting on block 2 are its weight (mg) and the normal force (Fn) from block 3.
The weight of block 2 is given by:
Weight2 = m2 * g
Where m2 is the mass of block 2 and g is the acceleration due to gravity (approximately 9.8 m/s^2).
The normal force between blocks 2 and 3 is equal in magnitude but opposite in direction to the weight of block 2. Therefore:
Fn = Weight2 = m2 * g
Plugging in the values we know:
Fn = (31.515 kg) * (9.8 m/s^2)
Fn ≈ 308.057 N (rounded to three decimal places)
Therefore, the normal force between blocks 2 and 3 is approximately 308.057 N.