A block and a triangular wedge move together

Problem20_2A block of mass m and a triangular wedge of mass M move together without slipping due to a pulling force F. The triangular wedge has a slope α. There is no friction between any pair of surfaces. Derive an equation of pulling force.


The block and the triangular wedge may move together without slipping only if they have the same acceleration. The acceleration will be in the same direction in which the pulling force F acts.

a = F/(M+m)

We draw a free-body diagram of the block.solProblem20_2

We apply Newton’s second law of motion

Fnet = ma

Fnet,x and Fnet,y are the components of the net force along the x and y axes respectively. The block does not accelerate along y axis, Fnet,y = may becomes Fnet,y = 0 or

Fy + FN,y – Fg = 0

In the x direction, Fnet,x = max becomes

Fx − FN,x = ma

We can rewrite the last two equations using the following facts:

Fy = F∙sin⁡(α), Fx = F∙cos⁡(α),
FN,y = FN∙cos⁡(α), FN,x = FN∙sin⁡(α)
and Fg = mg

and we get a system of two equations

F∙sin(α) + FN∙cos(α) − mg = 0
F∙cos(α) − FN∙sin(α) = ma

Also, we know that

a = F/(m+M)

and we rewrite the system of two equations using it

F∙sin(α) + FN∙cos(α) − mg = 0   (1)
F∙cos(α) − FN∙sin(α) = mF/(m+M)   (2)

We solve the system of two equations.
We divide the equation (1) by cos(α) and we get

FN = (mg − F∙sin(α)) / cos(α)   (3)

We substitute the equation (3) into the equation (2)

F∙cos(α) − (mg − F∙sin(α))∙sin(α)/cos(α) = mF/(m+M)

and multiply by cos(α)

F∙cos2(α) − mg∙sin(α) + F∙sin2(α) = m∙F∙cos(α)/(m+M)

We use the fact that

sin2(α) + cos2(α) = 1

and we can rewrite the last expression as

F − m∙F∙cos(α)/(m + M) = mg∙sin(α)

We take a common factor F out of the two terms in the left part of equation

F(1 − m∙cos(α)/(m + M)) = mg∙sin(α)

Finally, we divide the last equation by (1 − m∙cos(α)/(m + M)) and we get

F = mg∙sin(α)/ (1 − m∙cos(α)/(m + M))



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