A block of mass M1 = 288.0 kg sits on an inclined plane and is connected to a bucket with a massless string over a massless and
frictionless pulley. The coefficient of static friction between the block and the plane is μs = 0.41, and the angle between the plane and the horizontal is θ = 41°. Mass M2 can be changed by adding or taking away sand from the bucket. What is the maximum value of M2 for which the system will remain at rest?
1 answer:
Answer:
M2 = 278.06 kg
Explanation:
We calculate the weight of M1
W=m*g
Where
m: mass (kg)
g: acceleration due to gravity (m/s²)
W₁=288* 9.8= 2822.4 N
Look at the attached graphic
We calculate the x-y components of the weight :
W₁x= 2822.4*sin41° N =1851.66 N
W₁y= 2822.4 *cos41° N = 2130.09 N
We apply Newton's first law for the balance in M1:
Σ Fy=0
Fn-W₁y=0 , Fn: normal force
Fn=W₁y=2130.09N
Friction Force = Ff=μs *Fn = 0.41*2130.09 =873.34 N
Σ Fx=0
T- W₁x- Ff=0
T= 1851.66 + 873.34
T= 1851.66 + 873.34
T=2725 N
We apply Newton's first law for the balance in M2:
Σ Fy=0
T- W₂ =0
W₂ = T = 2725 N
W₂ = M2*g
M2 = W₂/g
M2 = 2725/9.8
M2 = 278.06 kg
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