Malcolm’s dad decides teaching high school chemistry just isn’t enough to make e

Question

Malcolm’s dad decides teaching high school chemistry just isn’t enough to make ends meet. He and a former student start a novel business enterprise. At one point, they need to move a large box of supplies of mass M up a ramp of angle =50°. The box is initially stationary, with a coefficient of static friction of s=0.75.

a. Walter has an idea. Attach the supplies to a box hanging over a pulley, of mass m. By adding weight to the box, Walter can change the amount of mass m. At what mass m will the large box just start to move? Solve in terms of variables.

b. If M = 30 kg, what is m? What forces are acting on mass M?

c. Jesse wasn’t paying attention, and doubles the m found in part b. What is the acceleration of the supplies box (mass M) if k=0.45?

Explanation / Answer

part a:

when the system is just about to move, the velocity of each block=0

hence tension in the string=T=m*g...(1)

also the tension should also balance the force on the bigger box of mass M along the ramp
hence T=M*g*sin(theta)+friction force

=M*g*sin(theta)+mu_s*M*g*cos(theta)

=M*g*(sin(theta)+mu_s*cos(theta))...(2)

combining equation 1 and 2,

m*g=M*g*(sin(theta)+mu_s*cos(theta))

==>m=M*(sin(theta)+mu_s*cos(theta))

part b:

given M=30 kg

theta=50 degree

mu_s=0.75

using these values, we get m=37.444 kg.

forces acting on M are:

weight of M along the inclince=M*g*sin(theta)

friction force , along the inclince, opposing the motion=mu_s*M*g*cos(theta)

tension in the cable=T

part c:

new value of m=2*37.444=74.888 kg

let acceleration of the system is a m/s^2

then writing force balance equation for m:

m*g-T=m*a...(3)

writing force balance equation for M:

T-M*g*sin(theta)-mu_k*M*g*cos(theta)=M*a...(4)

adding equation 3 and 4,

g*(m-M*sin(theta)-M*mu_k*cos(theta))=(m+M)*a

==>a=g*(m-M*sin(theta)-M*mu_k*cos(theta))/(m+M)

using the values of different variables,

acceleration=a=4.04 m/s^2

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