Homework 4 Begin Date: 9/27/2017 2:30:00 PM- Due Date: 10/9/2017 11:59:00 PM End

Question

Homework 4 Begin Date: 9/27/2017 2:30:00 PM- Due Date: 10/9/2017 11:59:00 PM End Date: 10/10/2017 4:00:00 ANM (8%) Problem 12: A block of mass m = 19.5 kg rests on an inclined plane with a coefficient of static friction of = 0.085 between the block and the plane. The inclined plane is L 6.8 mn long and it has a height of h 3.05 m at its tallest point. ©t heexpert ta.com 20% Part (a) what angle, in degrees, does the plane make with respect to the horizontal? Grade Summary Deductions 100% Potential 0% sinO | cos() | tan() | | (I ) 71819 4 5 6 Submissions cotan0 asinOacos0 atan) acotanosinh0 cosh) tanh0 cotanhO ODegrees Radians (5% per attempt) detailed view SubmitHint I give up! Hints: 5% deduction per hint. Hints remaining: 2 Feedback: 2% deduction per feedback. 20% Part (b) what is the magnitude of the normal force, FN in newtons, that acts on the block? 20% Part (c) what is the component of the force of gravity along the plane, Fer in newtons? 20% Part (d) write an expression, in terms of 0, the mass m the coefficient of static friction f's and the gravitational constant g, for the magnitude of the force due to static friction, F just before the block begins to slide. 20% Part (e) Will the block slide?

Explanation / Answer

Part (a)

h = 3.05 m, L = 6.8 m

Now, sin theta = h/L = 3.05/6.8 = 0.45

so, theta = sin^-1 (0.45) = 26.7 deg.

Part (b)

Normal Force, F(N) = m*g*cos theta = 19.5*9.81*cos26.7 = 170.9 N

Part (c)

Force of gravity acting along the inclined plane = m*g*sin theta = 19.5*9.81*sin26.7 = 85.9 N

Part (d)

Magnitude of force due to static friction, Fs = mu*m*g*cos theta

Part (e)

Value of Fs = 0.085*170.9 = 14.53 N

Since Fs < force of gravity acting along the inclined plane, So, the block will slide.

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