In the Figure, a block m 1 sits on a table. There is static friction between blo
ID: 1383787 • Letter: I
Question
In the Figure, a block m1sits on a table. There is static friction between block and table. Block m2hangs from a knot, as shown. Call the tension in the rope connecting the knot to m1, "T1". Call the tension in the rope connecting the knot to m2, "T2". Call the tension in the third rope (the one tipped up by an angle theta, connecting the knot to the wall), "T3". The system is in equilibrium.
In the previous problem, block m1 weighs 724 N. The coefficient of static friction between the block and the table is 0.28 and the angle theta is 25.0o. Find the maximum weight of block m2for which block m1 will remain at rest.
Explanation / Answer
due to weight of m2, the string attached to M2 will try to pull M1
now as the system is in equilibrium, the friction force on M1 is balancing the tension in the string connected to M1.
A)no,
as friction is the force that is preventing the block M1 from being pulled towards right and falling down.
B)as the system is in equilibrium, net force is 0.
c)as system is equilibrium, and the string connecting block M2 is vertical,
the weight of M2 is balanced b tension in string i.e. T2 and vertical component of T3.
hence T2 is lesser than m2*g.
d)force of static friction is balanced by T1.
hence it is equal to T1.
e)the horizontal component of T3 balances T1.
hence T3*cos(theta)=T1
as cos(theta) is always less than 1, T3 is greater than T1
so answers are:
FFFFT